<s>[INST] 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.
 [/INST] $$\sum_{cyc}\dfrac{1}{(b+c)^2+a^2} \le \dfrac{2}{25}(a+b+c)+\dfrac{9}{25}=\dfrac{3}{5}$$
$$\dfrac{1}{(b+c)^2+a^2}=\dfrac{1}{(3-a)^2+a^2}\le \dfrac{2}{25}a+\dfrac{3}{25}\ ,  (a\ge 0)$$
</s>