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!