MHB Can Trigonometric Inequalities Be Proven with Simple Equations?

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The discussion centers on proving the inequality $$\tan x+\tan y+\tan z\ge \sin x \sec y+\sin y\sec z+\sin z \sec x$$ for angles $x, y, z$ within the interval $(0, \frac{\pi}{2})$. Participants share their approaches and solutions, with one user expressing gratitude for contributions. The focus is on the methods used to demonstrate the inequality effectively. The conversation highlights the complexity of trigonometric inequalities and the collaborative effort to find a solution.
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Prove $$\tan x+\tan y+\tan z\ge \sin x \sec y+\sin y\sec z+\sin z \sec x$$ for $x,\,y,\,z\in \left(0,\,\dfrac{\pi}{2}\right)$.
 
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My solution:
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.
 
Bravo, lfdahl and thanks for participating!(Cool)