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lơn hơn 2 chứ Câu hỏi của Michelle Nguyen - Toán lớp 9 - Học toán với OnlineMath
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\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{\left(xy+yz+zx\right)^2}{x^2y^2z^2}\)(1) với x+y+z=0. Bạn quy đồng vế trái (1) dc \(\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}=\frac{\left(xy+yz+zx\right)^2-2\left(x+y+z\right)xyz}{x^2y^2z^2}\)
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Bớt 6 ở hai vế BĐT cần chứng minh tương đương:
\(\frac{\left(8c-a-b\right)\left(a-b\right)^2+\left(a+b\right)\left(a+b-2c\right)^2}{4abc}\le\frac{\left(7a+7b-2c\right)\left(a-b\right)^2+\left(a+b+2c\right)\left(a+b-2c\right)^2}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow\frac{1}{2}\left(a-b\right)^2\left[\frac{7a+7b-2c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{8c-a-b}{2abc}\right]+\frac{1}{2}\left(a+b-2c\right)^2\left[\frac{a+b+2c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{a+b}{2abc}\right]\ge0\)
Tới phần khó chừa lại cho bạn:V
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\(\frac{a^2}{b}-a+b+b=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\)
\(=\sqrt{a^2-ab+b^2}+\sqrt{a^2-ab+b^2}=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\)
\(\ge\sqrt{a^2-ab+b^2}+\sqrt{\frac{1}{4}\left(a+b\right)^2}=\sqrt{a^2-ab+b^2}+\frac{a+b}{2}\)
chứng minh tương tự ta được
\(\frac{b^2}{c}-b+c+c\ge\sqrt{b^2-bc+c^2}+\frac{b+c}{2},\frac{c^2}{a}-c+a+a\ge\sqrt{c^2-ca+a^2}+\frac{a+c}{2}\)
cộng vế với vế ta được
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}+a+b+c\)
Dấu bằng xảy ra khi a=b=c
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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)=9\left(dpcm\right)\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}.\text{ÁP DỤNG BĐT CÔ SI TA ĐƯỢC:}\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge3+2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{bc}{bc}}+2\sqrt{\frac{c}{a}.\frac{a}{c}}=3+2+2+2=9\)
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\(\frac{ax+by+cz}{xy}=z\Rightarrow z=\frac{a}{y}+\frac{b}{x}+\frac{cz}{xy}>\frac{a}{y}+\frac{b}{x}\)
Tương tự có \(y>\frac{a}{z}+\frac{c}{x}\); \(x>\frac{b}{z}+\frac{c}{y}\)
\(\Rightarrow x+y+z>\frac{b+c}{x}+\frac{a+c}{y}+\frac{a+b}{z}=\frac{b+c}{x}+x+\frac{a+c}{y}+y+\frac{a+b}{z}+z-x-y-z\)
\(\Rightarrow2\left(x+y+z\right)>2\sqrt{b+c}+2\sqrt{a+c}+2\sqrt{a+b}\)
\(\Rightarrow x+y+z>\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)