Bất đẳng thức

Question

Cho $a,b,c>0$.
Chứng minh rằng: $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+6\ge\frac{3}{2}(\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{a^2+c^2}{ac})$

score 4 · Answer 1

Bình chọn tăng 4 Bình chọn giảm

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