$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$ Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$
 A: You labeled this as homework, but I am not sure what methods you have learnt. (Are you in HSGS of Vietnam for example?) I would assume that you know the tangent line method, and will add more explanation if you need it.
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
$$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5} - \frac{18}{25}(a-1+b-1+c-1) = \frac{3}{5}$$
A: A full expanding gives
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)\geq0,$$
which is true by Schur and Muirhead:
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)=$$
$$=\sum_{cyc}(a^6-a^3b^3)+2\sum_{cyc}(a^6-a^2b^2c^2)+\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2)+$$
$$+3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+3\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Done!
By the way, the last inequality is true for all reals $a$, $b$ and $c$.
