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.Tuy nhiên mik có thể chữa lại đề cho ae dễ đọc nha:
Cho a,b,c>0 và:
\(P=\frac{a^3}{a^2}+ab+b^2+\frac{b^3}{b^2}+bc+c^2+\frac{c^3}{c^2}+ac+a^2.\)
\(Q=\frac{b^3}{a^2}+ab+b^2+\frac{c^3}{b^2}+bc+c^2+\frac{a^3}{c^2}+ac+a^2.\)
Chứng minh rằng:P=Q.
Ta có:
$P=\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}$
$Q=\dfrac{b^3}{a^2+ab+b^2}+\dfrac{c^3}{b^2+bc+c^2}+\dfrac{a^3}{c^2+ca+a^2}$
Xét $P-Q$:
$P-Q=\dfrac{a^3-b^3}{a^2+ab+b^2}+\dfrac{b^3-c^3}{b^2+bc+c^2}+\dfrac{c^3-a^3}{c^2+ca+a^2}$
Vì $\dfrac{x^3-y^3}{x^2+xy+y^2}=x-y$ nên $P-Q=(a-b)+(b-c)+(c-a)$$=0$
Suy ra: $P=Q$
cho c^2 +2(ab -ac -bc ) =0 và b khác c, a+b khác 0. Chứng minh a^2 +(a-c)^2 /b^2+(b-c)^2 = a-c / b-c
\(a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2\)
\(\Rightarrow\left(a+b-c\right)^2=a^2+b^2\)
\(\Rightarrow\hept{\begin{cases}a^2=\left(a+b-c\right)^2-b^2=\left(a+b-c-b\right)\left(a+b-c+b\right)=\left(a-c\right)\left(a+2b-c\right)\\b^2=\left(a+b-c\right)^2-a^2=\left(a+b-c-a\right)\left(a+b-c+a\right)=\left(b-c\right)\left(2a+b-c\right)\end{cases}}\)
\(a^2+\left(a-c\right)^2=\left(a-c\right)\left(a+2b-c\right)+\left(a-c\right)^2\)
\(=\left(a-c\right)\left(a+2b-c+a-c\right)=2\left(a-c\right)\left(a+b-c\right)\)
\(b^2+\left(b-c\right)^2=\left(b-c\right)\left(2a+b-c\right)+\left(b-c\right)^2\)
\(=\left(b-c\right)\left(2a+b-c+b-c\right)=2\left(b-c\right)\left(a+b-c\right)\)
Vậy \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{2\left(a-c\right)\left(a+b+c\right)}{2\left(b-c\right)\left(a+b+c\right)}=\frac{a-c}{b-c}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
<=>\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
<=>\(ab+bc+ca=0\)
<=>\(\frac{ab+bc+ca}{abc}=0\)
<=> \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
<=>\(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
<=>\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c}^3\)
<=>\(\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=\frac{-1}{c}^3\)
<=>\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Ta có: \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{3abc}{abc}=3\)
Bài 5.
1. Chứng minh
$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$
Ta có:
$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$
$=\dfrac{2b(a+b)+a(a+b)-4ab}{ab(a+b)}$
$=\dfrac{a^2-ab+2b^2}{ab(a+b)}$
$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$
Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$
2. Chứng minh
$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$
Vì $a,b,c>0$ nên:
$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$
Mà: $\dfrac1a>\dfrac{a}{a+b+c}$
Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$
1.
$a^3+b^4-ab(a+b)$
$=a^3+b^4-a^2b-ab^2$
$=a^2(a-b)+b^2(b-a)$
$=(a-b)(a^2-b^2)$
$=(a-b)^2(a+b)\ge0$
Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$
2.
$a^4+b^4-ab(a^2+b^2)$
$=a^4+b^4-a^3b-ab^3$
$=a^3(a-b)+b^3(b-a)$
$=(a-b)(a^3-b^3)$
$=(a-b)^2(a^2+ab+b^2)\ge0$
Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$
3.
$a^5+b^5-ab(a^3+b^3)$
$=a^5+b^5-a^4b-ab^4$
$=a^4(a-b)+b^4(b-a)$
$=(a-b)(a^4-b^4)$
$=(a-b)^2(a+b)(a^2+b^2)\ge0$
Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$
Ta có \(\frac{bc}{a^2}+\frac{ab}{c^2}+\frac{ac}{b^2}=\frac{\left(bc\right)^3+\left(ab\right)^3+\left(ac\right)^3}{\left(abc\right)^2}\)
Ta lại có (a+b+c)2=a2+b2+c2
=>a2+b2+c2+2(ab+bc+ac)= a2+b2+c2
=> 2(ab+bc+ac)=0=> ab+bc+ac=0
Ta cần chứng minh bài toán phụ x+y+z=0 thì
x3+y3+z3=3xyz
Ta thấy x+y+z=0=> x+y=-z
=> (x+y)3=-z3 => x3+3xy(x+y)+y3=-z3
=> x3+y3+z3=-3xy(x+y)=-3xy.(-z)=3xyz
Áp dụng vào bài toán ta có
ab+bc+ac=0 => (ab)3+(bc)3+(ac)3=3(abc)2
=> \(\frac{bc}{a^2}+\frac{ab}{c^2}+\frac{ac}{b^2}=\frac{3\left(abc\right)^2}{\left(abc\right)^2}=3\)
=> đpcm