Nguyễn Thị Thảo Linh
Giới thiệu về bản thân
"Văn chương không phải là liều thuốc an thần ru con người ta vào giấc ngủ uể oải mà là sắt nung bỏng rát thức tỉnh con người."
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2026-01-08 10:47:25
Gì vậy
2026-01-08 10:47:16
Đừng nói nhau đây là cộng đồng hỏi đáp ko phải để chửi
2026-01-08 10:46:41
Là người nhưng mình ko biết
2026-01-08 10:46:29
Đừng khịa nhau
2026-01-08 10:46:09
Ê nha đừng khịa nhau
2026-01-08 10:45:19
Chứng minh bất đẳng thức Step 1: Biến đổi bất đẳng thức cần chứng minh Bất đẳng thức cần chứng minh là (1+a2)(1+b2)(1+c2)−(1+abc)3≥0open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed is greater than or equal to 0(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3≥0, tương đương với (1+a2)(1+b2)(1+c2)≥(1+abc)3open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren cubed(1+𝑎2)(1+𝑏2)(1+𝑐2)≥(1+𝑎𝑏𝑐)3. Step 2: Áp dụng bất đẳng thức Cauchy-Schwarz Áp dụng bất đẳng thức Cauchy-Schwarz cho hai bộ số (1,a)open paren 1 comma a close paren(1,𝑎)và (1,b)open paren 1 comma b close paren(1,𝑏), ta có: (1+a2)(1+b2)≥(1⋅1+a⋅b)2=(1+ab)2open paren 1 plus a squared close paren open paren 1 plus b squared close paren is greater than or equal to open paren 1 center dot 1 plus a center dot b close paren squared equals open paren 1 plus a b close paren squared(1+𝑎2)(1+𝑏2)≥(1⋅1+𝑎⋅𝑏)2=(1+𝑎𝑏)2 Dấu bằng xảy ra khi a/1=b/1a / 1 equals b / 1𝑎/1=𝑏/1, tức là a=ba equals b𝑎=𝑏.
Khi đó, bất đẳng thức ban đầu trở thành: (1+ab)2(1+c2)≥(1+abc)3open paren 1 plus a b close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren cubed(1+𝑎𝑏)2(1+𝑐2)≥(1+𝑎𝑏𝑐)3 Step 3: Đặt ẩn phụ và chứng minh Đặt ab=xa b equals x𝑎𝑏=𝑥. Bất đẳng thức trở thành (1+x)2(1+c2)≥(1+xc)3open paren 1 plus x close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus x c close paren cubed(1+𝑥)2(1+𝑐2)≥(1+𝑥𝑐)3.
Khai triển hai vế: (1+2x+x2)(1+c2)≥1+3xc+3(xc)2+(xc)3open paren 1 plus 2 x plus x squared close paren open paren 1 plus c squared close paren is greater than or equal to 1 plus 3 x c plus 3 open paren x c close paren squared plus open paren x c close paren cubed(1+2𝑥+𝑥2)(1+𝑐2)≥1+3𝑥𝑐+3(𝑥𝑐)2+(𝑥𝑐)3 1+c2+2x+2xc2+x2+x2c2≥1+3xc+3x2c2+x3c31 plus c squared plus 2 x plus 2 x c squared plus x squared plus x squared c squared is greater than or equal to 1 plus 3 x c plus 3 x squared c squared plus x cubed c cubed1+𝑐2+2𝑥+2𝑥𝑐2+𝑥2+𝑥2𝑐2≥1+3𝑥𝑐+3𝑥2𝑐2+𝑥3𝑐3 c2+2x+2xc2+x2+x2c2−3xc−3x2c2−x3c3≥0c squared plus 2 x plus 2 x c squared plus x squared plus x squared c squared minus 3 x c minus 3 x squared c squared minus x cubed c cubed is greater than or equal to 0𝑐2+2𝑥+2𝑥𝑐2+𝑥2+𝑥2𝑐2−3𝑥𝑐−3𝑥2𝑐2−𝑥3𝑐3≥0 c2+2x+2xc2+x2−3xc−2x2c2−x3c3≥0c squared plus 2 x plus 2 x c squared plus x squared minus 3 x c minus 2 x squared c squared minus x cubed c cubed is greater than or equal to 0𝑐2+2𝑥+2𝑥𝑐2+𝑥2−3𝑥𝑐−2𝑥2𝑐2−𝑥3𝑐3≥0 Nhóm các hạng tử: (x2−2x2c2+x2c2)+(c2+2xc2)+(2x−3xc)−x3c3≥0open paren x squared minus 2 x squared c squared plus x squared c squared close paren plus open paren c squared plus 2 x c squared close paren plus open paren 2 x minus 3 x c close paren minus x cubed c cubed is greater than or equal to 0(𝑥2−2𝑥2𝑐2+𝑥2𝑐2)+(𝑐2+2𝑥𝑐2)+(2𝑥−3𝑥𝑐)−𝑥3𝑐3≥0 Điều này có vẻ phức tạp. Ta thử một cách khác. Step 4: Sử dụng phương pháp khác Ta biết rằng (1+a2)(1+b2)=1+a2+b2+a2b2=(a+b)2+(1−ab)2open paren 1 plus a squared close paren open paren 1 plus b squared close paren equals 1 plus a squared plus b squared plus a squared b squared equals open paren a plus b close paren squared plus open paren 1 minus a b close paren squared(1+𝑎2)(1+𝑏2)=1+𝑎2+𝑏2+𝑎2𝑏2=(𝑎+𝑏)2+(1−𝑎𝑏)2.
Bất đẳng thức cần chứng minh có thể được viết lại là: (1+a2)(1+b2)(1+c2)−(1+abc)3≥0open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed is greater than or equal to 0(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3≥0 Sử dụng hằng đẳng thức Euler cho 3 số: (1+a2)(1+b2)(1+c2)=(1+a2+b2+c2+a2b2+b2c2+c2a2+a2b2c2)open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren equals open paren 1 plus a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared plus a squared b squared c squared close paren(1+𝑎2)(1+𝑏2)(1+𝑐2)=(1+𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2+𝑎2𝑏2𝑐2) (1+abc)3=1+3abc+3a2b2c2+a3b3c3open paren 1 plus a b c close paren cubed equals 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed(1+𝑎𝑏𝑐)3=1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 Thay vào bất đẳng thức: 1+a2+b2+c2+a2b2+b2c2+c2a2+a2b2c2−(1+3abc+3a2b2c2+a3b3c3)≥01 plus a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared plus a squared b squared c squared minus open paren 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed close paren is greater than or equal to 01+𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2+𝑎2𝑏2𝑐2−(1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3)≥0 a2+b2+c2+a2b2+b2c2+c2a2−3abc−2a2b2c2−a3b3c3≥0a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 Điều này vẫn phức tạp. Step 5: Sử dụng Bất đẳng thức Schur Bất đẳng thức Schur bậc 1 cho x,y,z≥0x comma y comma z is greater than or equal to 0𝑥,𝑦,𝑧≥0: x(x−y)(x−z)+y(y−x)(y−z)+z(z−x)(z−y)≥0x open paren x minus y close paren open paren x minus z close paren plus y open paren y minus x close paren open paren y minus z close paren plus z open paren z minus x close paren open paren z minus y close paren is greater than or equal to 0𝑥(𝑥−𝑦)(𝑥−𝑧)+𝑦(𝑦−𝑥)(𝑦−𝑧)+𝑧(𝑧−𝑥)(𝑧−𝑦)≥0 Tương đương với x3+y3+z3+3xyz≥xy(x+y)+yz(y+z)+zx(z+x)x cubed plus y cubed plus z cubed plus 3 x y z is greater than or equal to x y open paren x plus y close paren plus y z open paren y plus z close paren plus z x open paren z plus x close paren𝑥3+𝑦3+𝑧3+3𝑥𝑦𝑧≥𝑥𝑦(𝑥+𝑦)+𝑦𝑧(𝑦+𝑧)+𝑧𝑥(𝑧+𝑥).
Đặt a2=x,b2=y,c2=za squared equals x comma b squared equals y comma c squared equals z𝑎2=𝑥,𝑏2=𝑦,𝑐2=𝑧. Bất đẳng thức cần chứng minh không liên quan trực tiếp đến dạng Schur. Step 6: Chứng minh lại bằng phương pháp biến đổi Quay lại (1+ab)2(1+c2)≥(1+abc)3open paren 1 plus a b close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren cubed(1+𝑎𝑏)2(1+𝑐2)≥(1+𝑎𝑏𝑐)3. (1+2ab+a2b2)(1+c2)≥1+3abc+3a2b2c2+a3b3c3open paren 1 plus 2 a b plus a squared b squared close paren open paren 1 plus c squared close paren is greater than or equal to 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed(1+2𝑎𝑏+𝑎2𝑏2)(1+𝑐2)≥1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 1+c2+2ab+2abc2+a2b2+a2b2c2≥1+3abc+3a2b2c2+a3b3c31 plus c squared plus 2 a b plus 2 a b c squared plus a squared b squared plus a squared b squared c squared is greater than or equal to 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed1+𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2+𝑎2𝑏2𝑐2≥1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 c2+2ab+2abc2+a2b2+a2b2c2−3abc−3a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared plus a squared b squared c squared minus 3 a b c minus 3 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2+𝑎2𝑏2𝑐2−3𝑎𝑏𝑐−3𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 c2+2ab+2abc2+a2b2−3abc−2a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 c2+2ab+2abc2+a2b2−3abc−2a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 Step 7: Sử dụng Bất đẳng thức Holder Bất đẳng thức Holder cho 3 dãy số (x1,y1,z1)open paren x sub 1 comma y sub 1 comma z sub 1 close paren(𝑥1,𝑦1,𝑧1), (x2,y2,z2)open paren x sub 2 comma y sub 2 comma z sub 2 close paren(𝑥2,𝑦2,𝑧2), (x3,y3,z3)open paren x sub 3 comma y sub 3 comma z sub 3 close paren(𝑥3,𝑦3,𝑧3):
(∑xip)1/p(∑yiq)1/q(∑zir)1/r≥∑xiyiziopen paren sum x sub i raised to the exponent p end-exponent close paren raised to the exponent 1 / p end-exponent open paren sum y sub i raised to the exponent q end-exponent close paren raised to the exponent 1 / q end-exponent open paren sum z sub i raised to the exponent r end-exponent close paren raised to the exponent 1 / r end-exponent is greater than or equal to sum x sub i y sub i z sub i(𝑥𝑝𝑖)1/𝑝(𝑦𝑞𝑖)1/𝑞(𝑧𝑟𝑖)1/𝑟≥𝑥𝑖𝑦𝑖𝑧𝑖với 1/p+1/q+1/r=11 / p plus 1 / q plus 1 / r equals 11/𝑝+1/𝑞+1/𝑟=1.
Không áp dụng được trực tiếp. Step 8: Chứng minh bất đẳng thức bằng cách xét hàm Đặt f(a,b,c)=(1+a2)(1+b2)(1+c2)−(1+abc)3f open paren a comma b comma c close paren equals open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed𝑓(𝑎,𝑏,𝑐)=(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3.
Ta cần chứng minh f(a,b,c)≥0f open paren a comma b comma c close paren is greater than or equal to 0𝑓(𝑎,𝑏,𝑐)≥0với a,b,c>0a comma b comma c is greater than 0𝑎,𝑏,𝑐>0.
Xét trường hợp a=b=ca equals b equals c𝑎=𝑏=𝑐. Khi đó: (1+a2)3−(1+a3)3=(1+3a2+3a4+a6)−(1+3a3+3a6+a9)open paren 1 plus a squared close paren cubed minus open paren 1 plus a cubed close paren cubed equals open paren 1 plus 3 a squared plus 3 a raised to the exponent 4 end-exponent plus a raised to the exponent 6 end-exponent close paren minus open paren 1 plus 3 a cubed plus 3 a raised to the exponent 6 end-exponent plus a raised to the exponent 9 end-exponent close paren(1+𝑎2)3−(1+𝑎3)3=(1+3𝑎2+3𝑎4+𝑎6)−(1+3𝑎3+3𝑎6+𝑎9) 3a2+3a4+a6−3a3−3a6−a9=3a2+3a4−3a3−a9=a2(3+3a2−3a−a7)3 a squared plus 3 a raised to the exponent 4 end-exponent plus a raised to the exponent 6 end-exponent minus 3 a cubed minus 3 a raised to the exponent 6 end-exponent minus a raised to the exponent 9 end-exponent equals 3 a squared plus 3 a raised to the exponent 4 end-exponent minus 3 a cubed minus a raised to the exponent 9 end-exponent equals a squared open paren 3 plus 3 a squared minus 3 a minus a raised to the exponent 7 end-exponent close paren3𝑎2+3𝑎4+𝑎6−3𝑎3−3𝑎6−𝑎9=3𝑎2+3𝑎4−3𝑎3−𝑎9=𝑎2(3+3𝑎2−3𝑎−𝑎7) Điều này không phải luôn dương. Có lẽ bất đẳng thức gốc sai hoặc điều kiện khác.
Kiểm tra lại đề bài. Đề bài là chứng minh (1+a2)(1+b2)(1+c2)−(1+abc)3≥0open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed is greater than or equal to 0(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3≥0.
Có thể bất đẳng thức là (1+a2)(1+b2)(1+c2)≥(1+abc)2open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren squared(1+𝑎2)(1+𝑏2)(1+𝑐2)≥(1+𝑎𝑏𝑐)2hoặc một dạng khác.
Giả sử đề bài đúng.
Khi a=b=c=1a equals b equals c equals 1𝑎=𝑏=𝑐=1, (1+12)3−(1+13)3=23−23=0open paren 1 plus 1 squared close paren cubed minus open paren 1 plus 1 cubed close paren cubed equals 2 cubed minus 2 cubed equals 0(1+12)3−(1+13)3=23−23=0.
Khi a=1,b=2,c=3a equals 1 comma b equals 2 comma c equals 3𝑎=1,𝑏=2,𝑐=3: (1+12)(1+22)(1+32)=2⋅5⋅10=100open paren 1 plus 1 squared close paren open paren 1 plus 2 squared close paren open paren 1 plus 3 squared close paren equals 2 center dot 5 center dot 10 equals 100(1+12)(1+22)(1+32)=2⋅5⋅10=100 (1+1⋅2⋅3)3=(1+6)3=73=343open paren 1 plus 1 center dot 2 center dot 3 close paren cubed equals open paren 1 plus 6 close paren cubed equals 7 cubed equals 343(1+1⋅2⋅3)3=(1+6)3=73=343 100−343<0100 minus 343 is less than 0100−343<0 Vậy bất đẳng thức như đề bài là sai.
Khi đó, bất đẳng thức ban đầu trở thành: (1+ab)2(1+c2)≥(1+abc)3open paren 1 plus a b close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren cubed(1+𝑎𝑏)2(1+𝑐2)≥(1+𝑎𝑏𝑐)3 Step 3: Đặt ẩn phụ và chứng minh Đặt ab=xa b equals x𝑎𝑏=𝑥. Bất đẳng thức trở thành (1+x)2(1+c2)≥(1+xc)3open paren 1 plus x close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus x c close paren cubed(1+𝑥)2(1+𝑐2)≥(1+𝑥𝑐)3.
Khai triển hai vế: (1+2x+x2)(1+c2)≥1+3xc+3(xc)2+(xc)3open paren 1 plus 2 x plus x squared close paren open paren 1 plus c squared close paren is greater than or equal to 1 plus 3 x c plus 3 open paren x c close paren squared plus open paren x c close paren cubed(1+2𝑥+𝑥2)(1+𝑐2)≥1+3𝑥𝑐+3(𝑥𝑐)2+(𝑥𝑐)3 1+c2+2x+2xc2+x2+x2c2≥1+3xc+3x2c2+x3c31 plus c squared plus 2 x plus 2 x c squared plus x squared plus x squared c squared is greater than or equal to 1 plus 3 x c plus 3 x squared c squared plus x cubed c cubed1+𝑐2+2𝑥+2𝑥𝑐2+𝑥2+𝑥2𝑐2≥1+3𝑥𝑐+3𝑥2𝑐2+𝑥3𝑐3 c2+2x+2xc2+x2+x2c2−3xc−3x2c2−x3c3≥0c squared plus 2 x plus 2 x c squared plus x squared plus x squared c squared minus 3 x c minus 3 x squared c squared minus x cubed c cubed is greater than or equal to 0𝑐2+2𝑥+2𝑥𝑐2+𝑥2+𝑥2𝑐2−3𝑥𝑐−3𝑥2𝑐2−𝑥3𝑐3≥0 c2+2x+2xc2+x2−3xc−2x2c2−x3c3≥0c squared plus 2 x plus 2 x c squared plus x squared minus 3 x c minus 2 x squared c squared minus x cubed c cubed is greater than or equal to 0𝑐2+2𝑥+2𝑥𝑐2+𝑥2−3𝑥𝑐−2𝑥2𝑐2−𝑥3𝑐3≥0 Nhóm các hạng tử: (x2−2x2c2+x2c2)+(c2+2xc2)+(2x−3xc)−x3c3≥0open paren x squared minus 2 x squared c squared plus x squared c squared close paren plus open paren c squared plus 2 x c squared close paren plus open paren 2 x minus 3 x c close paren minus x cubed c cubed is greater than or equal to 0(𝑥2−2𝑥2𝑐2+𝑥2𝑐2)+(𝑐2+2𝑥𝑐2)+(2𝑥−3𝑥𝑐)−𝑥3𝑐3≥0 Điều này có vẻ phức tạp. Ta thử một cách khác. Step 4: Sử dụng phương pháp khác Ta biết rằng (1+a2)(1+b2)=1+a2+b2+a2b2=(a+b)2+(1−ab)2open paren 1 plus a squared close paren open paren 1 plus b squared close paren equals 1 plus a squared plus b squared plus a squared b squared equals open paren a plus b close paren squared plus open paren 1 minus a b close paren squared(1+𝑎2)(1+𝑏2)=1+𝑎2+𝑏2+𝑎2𝑏2=(𝑎+𝑏)2+(1−𝑎𝑏)2.
Bất đẳng thức cần chứng minh có thể được viết lại là: (1+a2)(1+b2)(1+c2)−(1+abc)3≥0open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed is greater than or equal to 0(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3≥0 Sử dụng hằng đẳng thức Euler cho 3 số: (1+a2)(1+b2)(1+c2)=(1+a2+b2+c2+a2b2+b2c2+c2a2+a2b2c2)open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren equals open paren 1 plus a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared plus a squared b squared c squared close paren(1+𝑎2)(1+𝑏2)(1+𝑐2)=(1+𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2+𝑎2𝑏2𝑐2) (1+abc)3=1+3abc+3a2b2c2+a3b3c3open paren 1 plus a b c close paren cubed equals 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed(1+𝑎𝑏𝑐)3=1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 Thay vào bất đẳng thức: 1+a2+b2+c2+a2b2+b2c2+c2a2+a2b2c2−(1+3abc+3a2b2c2+a3b3c3)≥01 plus a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared plus a squared b squared c squared minus open paren 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed close paren is greater than or equal to 01+𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2+𝑎2𝑏2𝑐2−(1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3)≥0 a2+b2+c2+a2b2+b2c2+c2a2−3abc−2a2b2c2−a3b3c3≥0a squared plus b squared plus c squared plus a squared b squared plus b squared c squared plus c squared a squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑎2+𝑏2+𝑐2+𝑎2𝑏2+𝑏2𝑐2+𝑐2𝑎2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 Điều này vẫn phức tạp. Step 5: Sử dụng Bất đẳng thức Schur Bất đẳng thức Schur bậc 1 cho x,y,z≥0x comma y comma z is greater than or equal to 0𝑥,𝑦,𝑧≥0: x(x−y)(x−z)+y(y−x)(y−z)+z(z−x)(z−y)≥0x open paren x minus y close paren open paren x minus z close paren plus y open paren y minus x close paren open paren y minus z close paren plus z open paren z minus x close paren open paren z minus y close paren is greater than or equal to 0𝑥(𝑥−𝑦)(𝑥−𝑧)+𝑦(𝑦−𝑥)(𝑦−𝑧)+𝑧(𝑧−𝑥)(𝑧−𝑦)≥0 Tương đương với x3+y3+z3+3xyz≥xy(x+y)+yz(y+z)+zx(z+x)x cubed plus y cubed plus z cubed plus 3 x y z is greater than or equal to x y open paren x plus y close paren plus y z open paren y plus z close paren plus z x open paren z plus x close paren𝑥3+𝑦3+𝑧3+3𝑥𝑦𝑧≥𝑥𝑦(𝑥+𝑦)+𝑦𝑧(𝑦+𝑧)+𝑧𝑥(𝑧+𝑥).
Đặt a2=x,b2=y,c2=za squared equals x comma b squared equals y comma c squared equals z𝑎2=𝑥,𝑏2=𝑦,𝑐2=𝑧. Bất đẳng thức cần chứng minh không liên quan trực tiếp đến dạng Schur. Step 6: Chứng minh lại bằng phương pháp biến đổi Quay lại (1+ab)2(1+c2)≥(1+abc)3open paren 1 plus a b close paren squared open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren cubed(1+𝑎𝑏)2(1+𝑐2)≥(1+𝑎𝑏𝑐)3. (1+2ab+a2b2)(1+c2)≥1+3abc+3a2b2c2+a3b3c3open paren 1 plus 2 a b plus a squared b squared close paren open paren 1 plus c squared close paren is greater than or equal to 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed(1+2𝑎𝑏+𝑎2𝑏2)(1+𝑐2)≥1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 1+c2+2ab+2abc2+a2b2+a2b2c2≥1+3abc+3a2b2c2+a3b3c31 plus c squared plus 2 a b plus 2 a b c squared plus a squared b squared plus a squared b squared c squared is greater than or equal to 1 plus 3 a b c plus 3 a squared b squared c squared plus a cubed b cubed c cubed1+𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2+𝑎2𝑏2𝑐2≥1+3𝑎𝑏𝑐+3𝑎2𝑏2𝑐2+𝑎3𝑏3𝑐3 c2+2ab+2abc2+a2b2+a2b2c2−3abc−3a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared plus a squared b squared c squared minus 3 a b c minus 3 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2+𝑎2𝑏2𝑐2−3𝑎𝑏𝑐−3𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 c2+2ab+2abc2+a2b2−3abc−2a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 c2+2ab+2abc2+a2b2−3abc−2a2b2c2−a3b3c3≥0c squared plus 2 a b plus 2 a b c squared plus a squared b squared minus 3 a b c minus 2 a squared b squared c squared minus a cubed b cubed c cubed is greater than or equal to 0𝑐2+2𝑎𝑏+2𝑎𝑏𝑐2+𝑎2𝑏2−3𝑎𝑏𝑐−2𝑎2𝑏2𝑐2−𝑎3𝑏3𝑐3≥0 Step 7: Sử dụng Bất đẳng thức Holder Bất đẳng thức Holder cho 3 dãy số (x1,y1,z1)open paren x sub 1 comma y sub 1 comma z sub 1 close paren(𝑥1,𝑦1,𝑧1), (x2,y2,z2)open paren x sub 2 comma y sub 2 comma z sub 2 close paren(𝑥2,𝑦2,𝑧2), (x3,y3,z3)open paren x sub 3 comma y sub 3 comma z sub 3 close paren(𝑥3,𝑦3,𝑧3):
(∑xip)1/p(∑yiq)1/q(∑zir)1/r≥∑xiyiziopen paren sum x sub i raised to the exponent p end-exponent close paren raised to the exponent 1 / p end-exponent open paren sum y sub i raised to the exponent q end-exponent close paren raised to the exponent 1 / q end-exponent open paren sum z sub i raised to the exponent r end-exponent close paren raised to the exponent 1 / r end-exponent is greater than or equal to sum x sub i y sub i z sub i(𝑥𝑝𝑖)1/𝑝(𝑦𝑞𝑖)1/𝑞(𝑧𝑟𝑖)1/𝑟≥𝑥𝑖𝑦𝑖𝑧𝑖với 1/p+1/q+1/r=11 / p plus 1 / q plus 1 / r equals 11/𝑝+1/𝑞+1/𝑟=1.
Không áp dụng được trực tiếp. Step 8: Chứng minh bất đẳng thức bằng cách xét hàm Đặt f(a,b,c)=(1+a2)(1+b2)(1+c2)−(1+abc)3f open paren a comma b comma c close paren equals open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed𝑓(𝑎,𝑏,𝑐)=(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3.
Ta cần chứng minh f(a,b,c)≥0f open paren a comma b comma c close paren is greater than or equal to 0𝑓(𝑎,𝑏,𝑐)≥0với a,b,c>0a comma b comma c is greater than 0𝑎,𝑏,𝑐>0.
Xét trường hợp a=b=ca equals b equals c𝑎=𝑏=𝑐. Khi đó: (1+a2)3−(1+a3)3=(1+3a2+3a4+a6)−(1+3a3+3a6+a9)open paren 1 plus a squared close paren cubed minus open paren 1 plus a cubed close paren cubed equals open paren 1 plus 3 a squared plus 3 a raised to the exponent 4 end-exponent plus a raised to the exponent 6 end-exponent close paren minus open paren 1 plus 3 a cubed plus 3 a raised to the exponent 6 end-exponent plus a raised to the exponent 9 end-exponent close paren(1+𝑎2)3−(1+𝑎3)3=(1+3𝑎2+3𝑎4+𝑎6)−(1+3𝑎3+3𝑎6+𝑎9) 3a2+3a4+a6−3a3−3a6−a9=3a2+3a4−3a3−a9=a2(3+3a2−3a−a7)3 a squared plus 3 a raised to the exponent 4 end-exponent plus a raised to the exponent 6 end-exponent minus 3 a cubed minus 3 a raised to the exponent 6 end-exponent minus a raised to the exponent 9 end-exponent equals 3 a squared plus 3 a raised to the exponent 4 end-exponent minus 3 a cubed minus a raised to the exponent 9 end-exponent equals a squared open paren 3 plus 3 a squared minus 3 a minus a raised to the exponent 7 end-exponent close paren3𝑎2+3𝑎4+𝑎6−3𝑎3−3𝑎6−𝑎9=3𝑎2+3𝑎4−3𝑎3−𝑎9=𝑎2(3+3𝑎2−3𝑎−𝑎7) Điều này không phải luôn dương. Có lẽ bất đẳng thức gốc sai hoặc điều kiện khác.
Kiểm tra lại đề bài. Đề bài là chứng minh (1+a2)(1+b2)(1+c2)−(1+abc)3≥0open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren minus open paren 1 plus a b c close paren cubed is greater than or equal to 0(1+𝑎2)(1+𝑏2)(1+𝑐2)−(1+𝑎𝑏𝑐)3≥0.
Có thể bất đẳng thức là (1+a2)(1+b2)(1+c2)≥(1+abc)2open paren 1 plus a squared close paren open paren 1 plus b squared close paren open paren 1 plus c squared close paren is greater than or equal to open paren 1 plus a b c close paren squared(1+𝑎2)(1+𝑏2)(1+𝑐2)≥(1+𝑎𝑏𝑐)2hoặc một dạng khác.
Giả sử đề bài đúng.
Khi a=b=c=1a equals b equals c equals 1𝑎=𝑏=𝑐=1, (1+12)3−(1+13)3=23−23=0open paren 1 plus 1 squared close paren cubed minus open paren 1 plus 1 cubed close paren cubed equals 2 cubed minus 2 cubed equals 0(1+12)3−(1+13)3=23−23=0.
Khi a=1,b=2,c=3a equals 1 comma b equals 2 comma c equals 3𝑎=1,𝑏=2,𝑐=3: (1+12)(1+22)(1+32)=2⋅5⋅10=100open paren 1 plus 1 squared close paren open paren 1 plus 2 squared close paren open paren 1 plus 3 squared close paren equals 2 center dot 5 center dot 10 equals 100(1+12)(1+22)(1+32)=2⋅5⋅10=100 (1+1⋅2⋅3)3=(1+6)3=73=343open paren 1 plus 1 center dot 2 center dot 3 close paren cubed equals open paren 1 plus 6 close paren cubed equals 7 cubed equals 343(1+1⋅2⋅3)3=(1+6)3=73=343 100−343<0100 minus 343 is less than 0100−343<0 Vậy bất đẳng thức như đề bài là sai.
2026-01-08 10:44:08
Các nước Đông Nam Á có thế mạnh phát triển ngành trồng lúa nước nhờ các điều kiện tự nhiên và xã hội thuận lợi.
- Điều kiện tự nhiên:
- Khí hậu: Mang tính chất nhiệt đới gió mùa, nóng ẩm, mưa nhiều, cung cấp đủ nước và nhiệt độ thích hợp cho cây lúa nước sinh trưởng và phát triển.
- Địa hình và đất đai: Có nhiều đồng bằng châu thổ rộng lớn, đất phù sa màu mỡ, rất thích hợp cho việc canh tác lúa nước.
- Nguồn nước: Mạng lưới sông ngòi dày đặc, cung cấp nguồn nước tưới tiêu dồi dào.
- Điều kiện kinh tế - xã hội:
- Nguồn lao động: Dồi dào, có kinh nghiệm lâu đời trong việc trồng lúa nước.
- Thị trường: Nhu cầu tiêu thụ gạo trong khu vực và trên thế giới lớn.
- Chính sách: Nhiều quốc gia có chính sách khuyến khích phát triển nông nghiệp, đặc biệt là cấy lúa
2026-01-08 10:43:26
Đáp án đúng là C. Đơn giản, còn mang tính chất quân sự
2026-01-08 10:42:15
Sao vậy
2026-01-07 17:24:42
Cảm ơn bạn nhé