Inequality with the side lengths of a triangle

**veileen** · March 27th 2011, 12:22 AM

a, b, c - the side lengths of a triangle
k - a positive real number

$\frac{a^{k}}{(b+c)^{2}}+\frac{b^{k}}{(a+c)^{2}}+\f rac{c^{k}}{(b+a)^{2}}\geq \frac{1}{2}(\frac{a^{k}}{a^{2}+bc}+\frac{b^{k}}{b^ {2}+ac}+\frac{c^{k}}{c^{2}+ba})$

I observed that: $(b+c)^{2}+(a+c)^{2}+(b+a)^{2}=2(a^{2}+bc+b^{2}+ac+ c^{2}+ba)$ , so CBS inequality seems a good idea, but what I tried to do didn't help me ._.

I want some indications. Thanks in advance.

**Wilmer** · March 27th 2011, 03:16 AM

I observed:

a =< b =< c < a+b (since a triangle)

Your equation:
Equality obtained if a = b = c , else greater than

**veileen** · March 27th 2011, 12:17 PM

I used Cebîșev's inequality (supposing that a<=b<=c) and I got this:

$\sum_{a, b, c}\frac{1}{(a+b)^2}\geq \frac{1}{2}\sum_{a, b, c}\frac{1}{a^2+bc}$

And no, even if it looks easy, I don't know how to prove it >.>

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