Cho $ t, i, l > 0 $ thỏa $t.i.l =1$. Chứng minh:
$\sum\frac{t+1}{3(3+t)^2}\ge\sum\frac{1}{1+t+ti}$
Edited by Oral31211999, 31-03-2013 - 22:35.
$\sum\frac{t+1}{3(3+t)^2}\ge\sum\frac{1}{1+t+ti}$
Started By Oral1020, 29-03-2013 - 13:40
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