Cho $a,b,c$ là các số thực dương. Chứng minh rằng $\frac{abc}{(a+b)(b+c)(c+a)}\leq \frac{(a+b)(a+b+2c)}{(3a+3b+2c)^{2}}\leq \frac{1}{8}$
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Posted 13-04-2018 - 17:50
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