cho các số dương x,y,z thỏa mãn x.y.z=1. Chứng minh rằng:
[tex]\frac{x^{2}y^{2}}{2x^{2}+y^{2}+3x^{2}y^{2}}+\frac{y^{2}z^{2}}{2y^{2}+z^{2}+3y^{2}z^{2}}+\frac{z^{2}x^{2}}{2z^{2}+x^{2}+3z^{2}x^{2}}\leq \frac{1}{2}[/tex]
Đặt $\dfrac1{x^2}=a;\dfrac1{y^2}=b;\dfrac1{z^2}=c$
$\Rightarrow \dfrac{x^2y^2}{2x^2+y^2+3x^2y^2}=\dfrac{1}{2b+a+3}=\dfrac{1}{(a+b)+(b+1)+2}\leq \dfrac{1}{2\sqrt{ab}+2\sqrt{b}+2}$
Tương tự ta có $\dfrac{y^2z^2}{2y^2+z^2+3y^2z^2}\leq \dfrac{1}{2\sqrt{bc}+2\sqrt{c}+2};\dfrac{z^2x^2}{2z^2+x^2+3z^2x^2}\leq \dfrac{1}{2\sqrt{ca}+2\sqrt{a}+2}$
$\Rightarrow VT\leq \dfrac12(\dfrac1{\sqrt{ab}+\sqrt{b}+1}+\dfrac1{\sqrt{bc}+\sqrt{c}+1}+\dfrac1{\sqrt{ca}+\sqrt{a}+1}) \ (1)$
Mà $\sqrt{abc}=1$ $\Rightarrow \dfrac{1}{\sqrt{ab}+\sqrt{b}+1}+\dfrac{1}{\sqrt{bc}+\sqrt{c}+1}+\dfrac{1}{\sqrt{ca}+\sqrt{a}+1}
\\=\dfrac{1}{\sqrt{ab}+\sqrt{b}+1}+\dfrac{1}{\dfrac{1}{\sqrt{a}}+\sqrt{c}+1}+\dfrac{1}{\dfrac{1}{\sqrt{b}}+\sqrt{a}+1}
\\=\dfrac{1}{\sqrt{ab}+\sqrt{b}+1}+\dfrac{\sqrt{ab}}{\sqrt{b}+1+\sqrt{ab}}+\dfrac{\sqrt{b}}{1+\sqrt{ab}+\sqrt{b}}=1 \ (2)$
Từ (1) và (2) => đpcm
Dấu '=' xảy ra khi $a=b=c=1\Leftrightarrow x=y=z=1$