MHB Can You Prove the Minimum Value of a Trig Expression Given Certain Conditions?

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The discussion centers on a mathematical problem involving three angles, a, b, and c, constrained by the condition that their sine values sum to one. The goal is to prove that the sum of the squares of their tangents is at least 3/8. Participants share their solutions, with MarkFL and lfdahl providing correct responses. The thread emphasizes the importance of following the guidelines for problem submissions. The mathematical proof and alternate solutions contribute to a deeper understanding of trigonometric inequalities.
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Here is this week's POTW:

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Let that $a,\, b,\,c$ be three angles with $0<a,\,b,\,c<90^\circ$ that satisfy $\sin a+\sin b+\sin c=1$.

Prove that $\tan^2 a+\tan^2 b+\tan^2 c \ge \dfrac{3}{8}$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution:

1. MarkFL
2. lfdahl

Solution from MarkFL:
By cyclic symmetry, we know equality occurs for:

$$a=b=c=\arcsin\left(\frac{1}{3}\right)$$

And so the objective function has its critical point at:

$$f\left(\arcsin\left(\frac{1}{3}\right),\arcsin\left(\frac{1}{3}\right),\arcsin\left(\frac{1}{3}\right)\right)=3\tan^2\left(\arcsin\left(\frac{1}{3}\right)\right)=\frac{3}{8}$$

Evaluating the objection function at another point on the constraint, we find:

$$f\left(\arcsin\left(\frac{1}{4}\right),\arcsin\left(\frac{1}{2}\right),\arcsin\left(\frac{1}{4}\right)\right)=2\tan^2\left(\arcsin\left(\frac{1}{4}\right)\right)+\tan^2\left(\arcsin\left(\frac{1}{2}\right)\right)=\frac{7}{15}>\frac{3}{8}$$

Hence:

$$f_{\min}=\frac{3}{8}$$

Shown as desired. :)
Alternate solution from lfdahl:
Since $0^{\circ}< a,b,c < 90^{\circ}$, we can assume WLOG that $\sin a \ge \sin b \ge \sin c > 0$.

Hence

\[\sin^2 a \ge \sin^2 b \ge \sin^2 c\] and \[\frac{1}{\cos^2 a} \geq \frac{1}{cos^2 b} \geq \frac{1}{\cos^2 c}.\]

Applying Chebyschev´s sum inequality:

\[3\left (\tan^2 a + \tan^2 b + \tan^2 c \right ) \geq \left (\sin^2a + \sin^2b +\sin^2c\right )\left ( \frac{1}{\cos^2 a} + \frac{1}{cos^2 b} + \frac{1}{\cos^2 c} \right )\]

Now, again from Chebyschev´s sum inequality and from the given condition:

$\sin a + \sin b + \sin c = 1$, we have: $\sin^2a + \sin^2b +\sin^2c \geq \frac{1}{3}$.

Thus our inequality becomes:

\[3\left ( \tan^2 a + \tan^2 b + \tan^2 c \right ) \geq \frac{1}{3}\left (3 + \tan^2 a + \tan^2 b + \tan^2 c \right )\]

- or \[\tan^2 a + \tan^2 b + \tan^2 c \geq \frac{3}{8}.\]
Equality happens at $$a=b=c=\arcsin\left(\frac{1}{3}\right)$$.
 
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