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

[anemone]

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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[anemone]

Congratulations to the following members for their correct solution:

1. MarkFL
2. lfdahl

Solution from MarkFL:

Spoiler

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:

Spoiler

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