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

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
$$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5} - \frac{18}{25}(a-1+b-1+c-1) = \frac{3}{5}$$
