Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$

Question

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.

Answer 1

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!

Answer 2

$$\sum_{cyc}\dfrac{1}{(b+c)^2+a^2} \le \dfrac{2}{25}(a+b+c)+\dfrac{9}{25}=\dfrac{3}{5}$$

$$\dfrac{1}{(b+c)^2+a^2}=\dfrac{1}{(3-a)^2+a^2}\le \dfrac{2}{25}a+\dfrac{3}{25}\ , (a\ge 0)$$