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)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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