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[Gabriella Parkinson]

Cho $a,b,c> 0$

và $\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\leq 1$

Chứng minh rằng: 

$ab^{2}c^{3}\leq \frac{1}{5^{6}}$

Edited by Gabriella Parkinson, 29-03-2018 - 16:09.

[Tea Coffee]

$\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\leq 1(*)$

$(*)=>\frac{1}{1+a}\geq \frac{2b}{b+1}+\frac{3c}{c+1}\geq 5\sqrt[5]{\frac{b^{2}c^{3}}{(1+b)^{2}(1+c)^{3}}}$

$(*)\frac{1}{1+b}\geq \frac{a}{1+a}+\frac{b}{1+b} +\frac{3c}{1+c}\geq 5\sqrt[5]{\frac{abc^{3}}{(1+a)(1+b)(1+c)^{3}}}$

$(*)=>\frac{1}{1+c}\geq \frac{a}{1+a}+\frac{2b}{1+b}+\frac{2c}{1+c}\geq 5\sqrt[5]{\frac{ab^{2}c^{2}}{(1+a)(1+b)^{2}(1+c)^{2}}}$

Nhân vế theo vế $=>ab^{2}c^{3}\leq \frac{1}{5^{6}}$