Inequality on tangent and secant function 
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$


let $\tan\alpha =a, \tan\beta =b, \tan\gamma =c, \tan\delta =d\implies a,b,c,d>0$
Also $\alpha+\beta+\gamma+\delta=180^{\circ}\implies \alpha+\beta=180^{\circ}-(\gamma+\delta)$
$\implies \tan (\alpha+\beta)=-\tan (\gamma+\delta)$
$\implies \frac{a+b}{1-ab}=-\frac{c+d}{1-cd}$
$\implies a+b+c+d=abc+abd+acd+bcd$   ___(i)
Now Consider $(a+b)(a+c)(a+d)=a^3+(b+c+d)a^2+(bc+bd+cd)a+bcd$
$\implies (a+b)(a+c)(a+d)=a^2(a+b+c+d)+(a+b+c+d)$  (Using (i))
$\implies \frac{a^2+1}{a+b}=\frac{(a+c)(a+d)}{a+b+c+d}$
$\implies \frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}=\frac{(a+c)(a+d)+(b+a)(b+d)+(c+a)(c+b)+(d+b)(d+c)}{a+b+c+d}$
$=a+b+c+d$___(ii)
Now by C-S inequality
$\left((a+b)+(b+c)+(c+d)+(d+a)\right)\left(\frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}\right)\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies 2(a+b+c+d)^2\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$
$\implies \sqrt2\left(\tan\alpha +\tan\beta +\tan\gamma +\tan\delta\right)\ge\sec\alpha +\sec\beta +\sec\gamma +\sec\delta$
