Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$ Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that  $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$
Things I have done so far:
$$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$
$$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\frac{1}{2(a-1)^2+7-2a}\leq \frac{1}{7-2a}$$
Then, I tried to use the UCT to solve this problem.
I created the new inequality:
$$\frac{1}{7-2a}\leq \frac{1}{5}+m.(a-1)(*)$$
with $0<a<3$. I needed to find "m" which make (*) always true.After that, I found $$m=\frac{2}{25}$$
However, 
$$(*)\Leftrightarrow \frac{1}{7-2a}\leq \frac{1}{5}+\frac{2}{25}.(a-1)\Leftrightarrow 4(a-1)^{2}\leq 0$$
which is wrong with any $$a\in \mathbb{R} $$
Can you show me what my mistake is?
I hope you can have "smart" way to solve this problem.
Sorry, I am not good at English.

Hint: Use the Tangent Line method.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{(3-a)^2+a^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or 
$$\sum_{cyc}\left(\frac{a^2-3a+2}{2a^2-6a+9}+\frac{1}{5}(a-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2(2a+1)}{2a^2-6a+9}\geq0$$ and we are done!
