If $\sum\limits_{\text{cyc}}x=\frac\pi2$, then $2\sqrt{\sum\limits_{\text{cyc}}\tan x}\le\sum\limits_{\text{cyc}}\frac{\sqrt{\tan x}}{\cos x}$ 
Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=\frac{\pi}{2}$. Then,
  $$2\sqrt{\tan x+\tan y+\tan z}\leq \frac{\sqrt{\tan x}}{\cos x}+\frac{\sqrt{\tan y}}{\cos y} +\frac{\sqrt{\tan z}}{\cos z}$$

My try:
I have tried to use the formula of a triangle (if we have $a+b+c=\pi$).
So I make the following substitution :
$$x=0.5a \qquad y=0.5b \qquad z=0.5c$$
And now the idea is to use the formula :
$$\cos(A)^2=\frac{a^2+b^2-c^2}{2ab}$$
But if we make this substitution :
$$u=\frac{a}{b} \qquad v=\frac{b}{c} \qquad w=\frac{c}{a}$$
We get :
$$\cos(A)^2=\frac{u^2+1-\frac{1}{v^2}}{2\frac{u}{v}}$$
Furthermore we know with the condition that we have :
$$\tan a \tan b \tan c=\tan a+\tan b+\tan c$$
And $$uvw=1$$
So, in conclusion, we have two conditions on one inequality, but I think it's hard to isolate each variable.

Let $\tan{x}=a$, $\tan{y}=b$ and $\tan{z}=c$.
Thus, $a$, $b$ and $c$ are positives, $ab+ac+bc=1$ and we need to prove that
$$\sum_{cyc}\sqrt{a(1+a^2)}\geq2\sqrt{a+b+c}.$$
Indeed, by C-S and Schur we obtain:
$$\sum_{cyc}\sqrt{a(1+a^2)}=\sum_{cyc}\sqrt{a(a^2+ab+ac+bc)}=\sum_{cyc}\sqrt{a(a+b)(a+c)}=$$
$$=\sum_{cyc}\sqrt{a^2(a+b+c)+abc}=\sqrt{\sum_{cyc}\left(a^2(a+b+c)+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(a^2(a+b+c)+abc+2(ab(a+b+c)+abc)\right)}=$$
$$=\sqrt{\sum_{cyc}(a^3+3a^2b+3a^2c+5abc)}\geq\sqrt{\sum_{cyc}(4a^2b+4a^2c+4abc)}=$$
$$=2\sqrt{(a+b+c)(ab+ac+bc)}=2\sqrt{a+b+c}$$ and we are done! 
