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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.

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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!

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