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-   -   $\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$ (http://forum.mathscope.org/showthread.php?t=49500)

dangthanhbn 14-03-2015 05:20 PM

$\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$
 
Cho $\left\{\begin{matrix} a,b,c\geq 0 & \\ a+b+c=3 & \end{matrix}\right.$. Chứng minh rằng:
$\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$


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