Can Trigonometric Inequalities Be Proven with Simple Equations?

[anemone]

Prove \(\displaystyle \tan x+\tan y+\tan z\ge \sin x \sec y+\sin y\sec z+\sin z \sec x\) for $x,\,y,\,z\in \left(0,\,\dfrac{\pi}{2}\right)$.

 

[lfdahl]

My solution:

Spoiler

WLOG let $x \le y \le z$ and $x,y,z \in \left ( 0,\frac{\pi }{2} \right )$.

Then $0 < sinx \le siny \le sinz < 1$, and $1 \le secx \le secy \le secz $.

Then, the result follows immediately from the Rearrangement Inequality:

\[tanx + tany + tanz = sinxsecx+sinysecy+sinzsecz \geq sinxsecy+sinysecz + sinzsecx\]

Any permutation of the RHS will obey the inequality.

 

[anemone]

Bravo, lfdahl and thanks for participating!(Cool)