1 reply to this topic

[killerdark68]

cho a,b,c >0 cmr $\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\geq a^3+b^3+c^3$

[Kira Tatsuya]

Áp dụng AM-GM:$\frac{a^5}{bc}+abc\geq 2\sqrt{\frac{a^5}{bc}.abc}\geq2\sqrt{a^6}\geq2a^3$

tương tự:

$\Rightarrow \frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\geq 2(a^3+b^3+c^3)-3abc\geq 2(a^3+b^3+c^3)-(a^3+b^3+c^3)\geq a^3+b^3+c^3$

Edited by Kira Tatsuya, 29-11-2015 - 14:57.