Inequality on tangent and secant function

Question

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

Answer 1

Let $f(x)=\sqrt2\tan{x}-\frac{1}{\cos{x}}$.

Thus, $$f''(x)=\frac{(1+\sqrt2-\sin{x})(\sin{x}-\sqrt2+1)}{\cos^3x}>0$$ for $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right)$.

Thus, by Vasc's RCF Theorem it's enough to prove our inequality for

$\beta=\gamma=\alpha$ and $\delta=\pi-3\alpha$, where $\alpha\in\left[\frac{\pi}{4},\frac{\pi}{3}\right)$ and the rest is smooth.

Indeed, we need to prove that $$\frac{3(\sqrt2\sin{\alpha}-1)}{\cos{\alpha}}+\frac{\sqrt{2}\sin\left(180^{\circ}-3\alpha\right)-1}{\cos\left(180^{\circ}-3\alpha\right)}\geq0.$$ Let $\sin\alpha=t$.

Thus, $\frac{1}{\sqrt2}\leq t<\frac{\sqrt3}{2}$ and we need to prove that $$\frac{3(\sqrt2t-1)}{\cos\alpha}-\frac{\sqrt2\sin3\alpha-1}{\cos3\alpha}\geq0$$ or $$3(\sqrt2t-1)-\frac{\sqrt2(3t-4t^3)-1}{4\cos^2\alpha-3}\geq0$$ or $$3(\sqrt2t-1)-\frac{\sqrt2(3t-4t^3)-1}{4(1-t^2)-3}\geq0$$ or $$3(\sqrt2t-1)+\frac{\sqrt2(3t-4t^3)-1}{4t^2-1}\geq0$$ or $$4\sqrt2t^3-6t^2+1\geq0,$$ which is true by AM-GM: $$4\sqrt2t^3+1=2\sqrt2t^3+2\sqrt2t^3+1\geq3\sqrt[3]{\left(2\sqrt2t^3\right)^2\cdot1}=6t^2.$$ Done!

We can use also the Tangent Line method. $$\sum_{cyc}\left(\sqrt2\tan\alpha-\sec\alpha\right)=\sum_{cyc}\left(\sqrt2\tan\alpha-\sec\alpha-\sqrt2\left(\alpha-\frac{\pi}{4}\right)\right)\geq0$$ because $f(x)\geq0$ for all $0<x<\frac{\pi}{2}$, where $$f(x)=\sqrt2\tan{x}-\sec{x}-\sqrt2\left(x-\frac{\pi}{4}\right).$$ Indeed, $$f'(x)=\frac{\sqrt2}{\cos^2x}-\frac{\sin{x}}{\cos^2x}-\sqrt2=\frac{\sin{x}(\sqrt2\sin{x}-1)}{\cos^2x},$$ which gives $x_{min}=\frac{\pi}{4}.$

Id est, $$f(x)\geq f\left(\frac{\pi}{4}\right)=0$$ and we are done!

Answer 2

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

$\left((a+b)+(b+c)+(c+d)+(d+a)\right)\left(\frac{a^2+1}{a+b}+\frac{b^2+1}{b+c}+\frac{c^2+1}{c+d}+\frac{d^2+1}{d+a}\right)\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$

$\implies 2(a+b+c+d)^2\ge\left(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\right)^2$

$\implies \sqrt2\left(\tan\alpha +\tan\beta +\tan\gamma +\tan\delta\right)\ge\sec\alpha +\sec\beta +\sec\gamma +\sec\delta$