Cho $x,y,z>0;  xyz=1$.Tìm giá trị lớn nhất của biểu thức:
$P=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}$.
Áp dụng bất đẳng thức Côsi, ta có: $\begin{cases}x^2+y^2\geq 2xy \\ y^2+1\geq 2y \end{cases}$
Từ đó suy ra: $x^2+2y^2+3\geq 2(xy+y+1)\Rightarrow \frac{1}{x^2+2y^2+3}\leq \frac{1}{2(xy+y+1)}  (1)$
Lập luận tương tự có: $\frac{1}{y^2+2z^2+3}\leq \frac{1}{2(yz+z+1)}    (2)$
                                        $\frac{1}{z^2+2x^2+3}\leq \frac{1}{2(zx+x+1)}    (3)$
Dấu bằng trong $(1),(2),(3)$ xảy ra khi và chỉ khi $x=y=z=1$.
Cộng từng vế với vế của  $(1),(2),(3)$ và có:
$P\leq \frac{1}{2}(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+ \frac{1}{zx+x+1})   (4)$
Dấu bằng trong $(4)$ xảy ra khi và chỉ khi $x=y=z=1$.
Vì  $xyz=1$ nên viết lại $(1)$ dưới dạng:
$P\leq \frac{1}{2}\left ( \frac{1}{xy+y+1}+\frac{xy}{xy^2z+xyz+xy}+ \frac{y}{xyz+xy+y} \right )   $

hay $P\leq \frac{1}{2}\left ( \frac{1}{xy+y+1}+\frac{xy}{xy+y+1}+ \frac{y}{xy+y+1} \right )   \Rightarrow P\leq \frac{1}{2}$.
Vậy $P_{ \max}=\frac{1}{2}\Leftrightarrow x=y=z=1$.
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