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Question

Cho

$a,b,c>0$ .
Chứng minh rằng:

$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+6\ge\frac{3}{2}(\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{a^2+c^2}{ac})$

khangnguyenthanh · Answer 1 · 10/20/2012 11:20:01 PM

Đặt:

$x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ , ta có:

$xyz=1$
Ta cần CM:

$x^2+y^2+z^2+6\ge\frac{3}{2}\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$
Đặt:

$f(x,y,z)=x^2+y^2+z^2+6-\frac{3}{2}\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$
Không mất tính tổng quát, giả sử:

$x=\min\{x,y,z\},t=\sqrt{yz}$ .
Ta có:

$f(x,y,z)-f(x,t,t)=\frac{1}{2}(\sqrt y-\sqrt z)^2\left(2(\sqrt y+\sqrt z)^2-3-\frac{3}{yz}\right)$

$\ge\frac{1}{2}(\sqrt y-\sqrt z)^2(8-3-3)\ge0$
Lại có:

$f(x,t,t)=f(\frac{1}{t^2},t,t)=\frac{(t-1)^2((t^2-2t-1)^2+t^2+1)}{2t^4}\ge0$ , đpcm.
Dấu bằng xảy ra khi:

$x=y=z=1\Leftrightarrow a=b=c$

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