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[cvp]

Cho $a;b;c$ là các số thực dương không âm và $abc=1$. Chứng minh rằng:
$\dfrac{1}{a^{2}+2b^{2}+3}+\dfrac{1}{b^{2}+2c^{2}+3}+\dfrac{1}{c^{2}+2a^{2}+3}\leq \dfrac{1}{2}$

Edited by Cao Xuân Huy, 28-12-2011 - 21:59.

[perfectstrong]

Lời giải:
\[\begin{array}{l}
{a^2} + 2{b^2} + 3 = {a^2} + {b^2} + {b^2} + 1 + 2 \ge 2ab + 2b + 2 \\
\Rightarrow \dfrac{1}{{{a^2} + 2{b^2} + 3}} \le \dfrac{1}{2}.\dfrac{1}{{ab + b + 1}} \\
\Rightarrow \dfrac{1}{{{a^2} + 2{b^2} + 3}} + \dfrac{1}{{{b^2} + 2{c^2} + 3}} + \dfrac{1}{{{c^2} + 2{a^2} + 3}} \le \dfrac{1}{2}\left( {\dfrac{1}{{ab + b + 1}} + \dfrac{1}{{bc + c + 1}} + \dfrac{1}{{ca + a + 1}}} \right) \\
= \dfrac{1}{2}\left( {\dfrac{1}{{ab + b + 1}} + \dfrac{1}{{\dfrac{1}{a} + \dfrac{1}{{ab}} + 1}} + \dfrac{1}{{\dfrac{1}{b} + a + 1}}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{{ab + b + 1}} + \dfrac{{ab}}{{b + 1 + ab}} + \dfrac{b}{{1 + ab + b}}} \right) = \dfrac{1}{2} \\
\end{array}\]