# $\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$

Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$

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Are you the same person? If so, I can flag the moderator to merge your accounts. Kindly use a single account. – user17762 Jan 20 '13 at 3:46

Since LHS is homogeneous, we may assume that $a+b+c = 3$. So we need to prove that $$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5}$$ Note that $$\frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} - \frac{18}{25}(a-1)$$ (This is equivalent to $(2a+1)(a-1)^2 \ge 0$) So $$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5} - \frac{18}{25}(a-1+b-1+c-1) = \frac{3}{5}$$
Neat. +1. Could you add more details for the step? $$\frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} - \frac{18}{25}(a-1)$$ – user17762 Jan 20 '13 at 4:02
@Marvis, You can check by clearing the denominator that it is equivalent to $(2a+1)(a-1)^2 \ge 0$. As for the motivation, let $f(a) = \frac{(3-2a)^2}{(3-a)^2+a^2}$. Then $f(1) = \frac{1}{5}$, and $f'(1) = -\frac{18}{25}$. – user27126 Jan 20 '13 at 4:04
Why you can assume $a+b+c=3$ ? – Toan Pham Quang Jan 20 '13 at 4:18
@harrypham, if not, replace $a$ by $\frac{3a}{a+b+c}$ and so forth. Left hand side doesn't change this way. – user27126 Jan 20 '13 at 4:19