$f(u)=1/(1+u^2)$ being convex, quantity is $\sum (2\cos^2-1)=2(\sum f(\tan))-3\ge 6f(\sum\tan/3)\ge 6f(1)-3=0$

dgrozev wrote: Suppose $\alpha,\beta,\gamma \in [0.\pi/2)$ and $\tan \alpha + \tan\beta + \tan \gamma \leq 3$. Prove that: \[\cos 2\alpha + \cos 2\beta + \cos 2\gamma \ge 0\] $\sum_{cyc}\cos 2\alpha =\sum_{cyc}(\frac{1-\tan^2\alpha}{1+\tan^2\alpha}+\tan\alpha-1)+3-\tan \alpha- \tan\beta -\tan \gamma= $ $=\sum_{cyc}\frac{\tan\alpha(\tan\alpha-1)^2}{1+\tan^2\alpha}+3-\tan \alpha- \tan\beta -\tan \gamma\geq0$.

We have that $cos 2\alpha=(1-x^2)/(1+x^2)$ where $x=tan\alpha$ and similar equations hold for the others. We have to prove that sum $(1-x^2)/(1+x^2)=>0$ (1) if $x, y, z$ are non-negative reals and $x+y+z<=3$. Add $3$ to both sides of (1) and devide by 2 to transform it to sum $1/(1+x^2)=>3/2$ (2). Now multiply it by $-1$ and add $3$ to both sides to transform it to sum $x^2/(1+x^2)<=3/2$ (3). Now obviously $1+x^2=>2x$ and we get that the $LHS$ of (3) is $<=(x+y+z)/2<=3/2$, which is what we wanted.