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123phandau


cm
[tex]\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a} + \frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \ge\ a + b + c [/tex] với a,b,c>0
[tex]\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a} + \frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \ge\ a + b + c [/tex] với a,b,c>0
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