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There have been 1 items by Gabriella Parkinson (Search limited from 22-05-2020)
Posted by Gabriella Parkinson on 29-03-2018 - 16:09 in Bất đẳng thức và cực trị
Cho $a,b,c> 0$
và $\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\leq 1$
Chứng minh rằng:
$ab^{2}c^{3}\leq \frac{1}{5^{6}}$