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ta có:\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
=\(\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(a+c\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
>= \(\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\)(BĐT Svaxo)=\(\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
>= \(\frac{3\sqrt[3]{a^2b^2c^2}}{2}\left(BĐTAM-GM\right)=\frac{3}{2}\)(đpcm)
dấu = khi a=b=c=1
ta có bđt phụ đã dc học
\(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) nếu bạn chưa học thì mik chứng mik cho:v
=> \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
=> \(3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2\ge0\)
=> \(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
điều này luôn đúng với mọi x;y;z
=>\(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
thay \(x=a+\frac{1}{a};y=b+\frac{1}{b};z=c+\frac{1}{c}\) vào ta có:
\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{\left(\left(a+b+c\right)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)^2}{3}\)
ta có bđt cosi mà thực ra mik cx ko nhớ tên nếu gọi việt mik thì gọi là bđt cộng mẫu nếu bạn ko bt mik lại chứng minh
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
ta nhân (a+b+c) vào hai vế:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
=\(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)
mà \(\frac{x}{y}+\frac{y}{x}\ge2\)
vì \(\frac{\left(x^2+y^2\right)}{xy}\ge2\)
\(x^2+y^2\ge2xy\)
=> \(\left(x^2-2xy+y^2\right)\ge0\) hay \(\left(x-y\right)^2\ge0\)
vậy x;y là các số thực thì \(\frac{x}{y}+\frac{y}{x}\ge2\)
=> 3+\(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge3+2+2+2=9\)
vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
thay vòa biểu thức đã suy ra ở đầu bài ta có:
=> \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{\left(\left(a+b+c\right)+\left(\frac{9}{a+b+c}\right)\right)^2}{3}\)
mà ta có a+b+c=1 thay vào biểu thức ta có:
\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{\left(1+9\right)^2}{3}=\frac{10^2}{3}=\frac{100}{3}\)
Ta có:
\(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\frac{1}{a^2}}{\frac{b+c}{bc}}\)
Biến đổi tương tự cho 2 BĐT còn lại ta có:
\(\frac{1}{b^3\left(a+c\right)}=\frac{\frac{1}{b^2}}{\frac{a+c}{ac}};\frac{1}{c^3\left(a+b\right)}=\frac{\frac{1}{c^2}}{\frac{a+b}{ab}}\)
Cộng theo vế 3 BĐT trên ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{2}{a}+\frac{2}{b}+\frac{2}{c}}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{1}{2}3\sqrt[3]{\left(abc\right)^2}=\frac{3}{2}\)
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)
Đặt vế trái là P
Ta có: \(P=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Ta có: abc = 1, thế vào ta được:
\(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\)
\(=\frac{b^2c^2}{a^2bc\left(b+c\right)}+\frac{c^2a^2}{b^2ac\left(c+a\right)}+\frac{a^2b^2}{c^2ab\left(a+b\right)}\)
Áp dụng BĐT Cauchy - Schwarz dạng Engel, ta có:
\(VT\ge\frac{\left(bc+ca+ac\right)^2}{abc\left(2ab+2bc+2ca\right)}=\frac{\left(bc+ca+ac\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
b) với mọi a,b,c ϵ R và x,y,z ≥ 0 có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(1\right)\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Thật vậy với a,b∈ R và x,y ≥ 0 ta có:
\(\frac{a^2}{x}=\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(2\right)\)
⇔\(\frac{a^2y}{xy}+\frac{b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}.\left(x+y\right)xy\ge\frac{\left(a+b\right)^2}{x+y}.\left(x+y\right)xy\)
⇔\(\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2xy\)
⇔\(a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+2abxy+b^2xy\)
⇔\(b^2x^2+a^2y^2-2abxy\ge0\)
⇔\(\left(bx-ay\right)^2\ge0\)(luôn đúng )
Áp dụng BĐT (2) có:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}
\)
= \(\frac{1}{a^2}.\frac{1}{ab+ac}+\frac{1}{b^2}.\frac{1}{bc+ac}+\frac{1}{c^2}.\frac{1}{ac+bc}\)
=\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)
Áp dụng BĐT (1) ta có:
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}++\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}\)
Mà abc=1⇒\(\left\{{}\begin{matrix}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{matrix}\right.\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\sqrt[3]{\frac{1}{1}}=3\)( BĐT cosi )
⇒\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
⇒\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}.3=\frac{3}{2}\)
Vậy \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Chúc bạn học tốt !!!
\(abc=1\ge a^2b^2c^2=1\)
\(\Rightarrow\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{b^2c^2}{a\left(b+c\right)}+\frac{a^2c^2}{b\left(c+a\right)}+\frac{a^2b^2}{c\left(a+b\right)}\)
Theo Cauchy-Schwarz ta được:
\(VP\ge\frac{\left(bc+ab+ac\right)^2}{2\left(ab+ac+bc\right)}=\frac{bc+ab+ac}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi a = b = c = 1
\(\)
Bài làm:
Ta có: \(\frac{1}{a^3\left(b+c\right)}=\frac{abc}{a^3\left(b+c\right)}=\frac{bc}{a^2b+a^2c}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}=\frac{b^2c^2}{ab+ac}\)
Tương tự: \(\frac{1}{b^3\left(c+a\right)}=\frac{c^2a^2}{ba+bc}\) ; \(\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2}{ca+cb}\)
=> \(Vt=\frac{a^2b^2}{ca+bc}+\frac{b^2c^2}{ab+ca}+\frac{c^2a^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
\(abc=1\Rightarrow\left(abc\right)^2=a^2b^2c^2=1\Rightarrow a^2=\frac{1}{b^2c^2}\Rightarrow\frac{1}{a^3\left(b+c\right)}=\frac{b^2c^2}{a\left(b+c\right)}=\frac{\left(bc\right)^2}{ab+ac}\)
Chứng minh tương tự ta có: \(\frac{1}{b^3\left(c+a\right)}=\frac{\left(ca\right)^2}{bc+ba};\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{ca+cb}\)
=> \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel: \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{bc+ca+ab+ca+ab+bc}=\frac{ab+bc+ca}{2}\)
Tiếp tục áp dụng bđt Cauchy với 3 số dương ta được: \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3\sqrt[3]{1}}{2}=\frac{3}{2}\)
=> \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{ab+bc+ca}{2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
viết lại sigma 1/a3(b+c) = sigma 1/a2/a(b+c)
đến đây dùng schwarz -> sigma 1/a3(b+c) >/ ab+bc+ac/2 >/ 3/2 (AM-GM cho ab,bc,ca)