1 reply to this topic

[RealCielo]

$\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\geq 3+\frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b+c)^2}$ voi a,b,c>0

Edited by RealCielo, 31-01-2016 - 16:19.

[quoccuonglqd]

Biến đổi tương đương

$\Leftrightarrow (a-b)^{2}\frac{a^{2}+b^{2}+ab+bc+ca}{(a+c)(b+c)(a+b+c)^{2}}+(b-c)^{2}\frac{b^{2}+c^{2}+ab+bc+ca}{(a+b)(a+c)(a+b+c)^{2}}+(c-a)^{2}\frac{c^{2}+a^{2}+ab+bc+ca}{(a+b)(b+c)(a+b+c)^{2}}\geqslant 0$