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

  • MHB
  • Thread starter anemone
  • Start date
  • Tags
    2017
  • #1
anemone
Gold Member
MHB
POTW Director
3,882
115
Here is this week's POTW:

-----

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

-----

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!
 
Physics news on Phys.org
  • #2
Congratulations to the following members for their correct solution:

1. MarkFL
2. lfdahl

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

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

And so the objective function has its critical point at:

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

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

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

Hence:

\(\displaystyle f_{\min}=\frac{3}{8}\)

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