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$\frac{abc}{(a+b)(b+c)(c+a)}\leq ...$

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hoangkimca2k2

hoangkimca2k2

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Cho $a,b,c$ là các số thực dương. Chứng minh rằng $\frac{abc}{(a+b)(b+c)(c+a)}\leq \frac{(a+b)(a+b+2c)}{(3a+3b+2c)^{2}}\leq \frac{1}{8}$


  N.D.P 





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