If $a^2+b^2+c^2+d^2+e^2=5$ so $\sum\limits_{cyc}\frac{1}{7-2a}\leq1$. 
Let $a$, $b$, $c$, $d$ and $d$ be non-negative numbers such that $a^2+b^2+c^2+d^2+e^2=5$. Prove that:
  $$\frac{1}{7-2a}+\frac{1}{7-2b}+\frac{1}{7-2c}+\frac{1}{7-2d}+\frac{1}{7-2e}\leq1$$

The equality occurs also for $a=2$ and $b=c=d=e=\frac{1}{2}$.
I tried TL:
$$1-\sum_{cyc}\frac{1}{7-2a}=\sum_{cyc}\left(\frac{1}{5}-\frac{1}{7-2a}\right)=\frac{2}{5}\sum_{cyc}\frac{1-a}{7-2a}=$$
$$=\frac{2}{5}\sum_{cyc}\left(\frac{1-a}{7-2a}+\frac{a^2-1}{10}\right)=\frac{1}{25}\sum_{cyc}\frac{(a-1)^2(3-2a)}{7-2a}$$
and I don't see what is the rest.

We must to prove that $f(a_1)+f(a_2)+f(a_3)+f(a_4)+f(a_5) \le 1$ , 
for non-negative $a_1+a_2+a_3+a_4+a_5=5$
$$f(x)=\dfrac{1}{7-2\sqrt{x}}$$
Since $f''(x)=\dfrac{(7-6\sqrt{x})}{2(2x-7\sqrt{x})^3}$ , we only need to consider the inequality in case $0< a_1=a_2=a_3=a_4=t^2 \le 1 \ , \ a_5=5-4t^2$
Clearly
$g(t)=\dfrac{4}{7-2t}+\dfrac{1}{7-2\sqrt{5-4t^2}} $
$ g'(t)=\dfrac{8}{(7-2t)^2}-\dfrac{8t}{\sqrt{5-4t^2}(7-2\sqrt{5-4t^2})^2}\ge 0 \Leftrightarrow $
$\sqrt{5-4t^2}\cdot(69-16t^2)  \ge 4t^3-140t^2+49t+140>0 \Leftrightarrow$
$(t-1)(2t-1)(t+1)(104t^3-60t^2+1062t-841) \le 0 \ ,\ t\in [0,1]$
So maximum of $g(t)$ is attained at $t_1=\dfrac{1}{2}$ and $t_2=1$. 
$$g(t_1)=g(t_2)=1$$
Equality holdes for : $(a=b=c=d=e=1) $ and $\left( a=b=c=d=\dfrac{1}{2}, e=2\right)$
