1
$\begingroup$

How would one show that, given $x,y,z,t\in[0,\pi]$, $$ \sin x+\sin y+\sin z+\sin t\leq4\sin\left(\frac{x+y+z+t}{4}\right) ? $$ I recognize that $\frac{x+y+z+t}{4}$ is the arithmetic mean of $x,y,z,t$ but I don't see how to prove the inequality. Expressing the $\sin$ as integrals of $\cos$ seems to lead nowhere... Just a hint would be appreciated.

$\endgroup$
1
$\begingroup$

You could start with proving $\sin x +\sin y\le 2\sin \left(\frac {x+y}2\right)$ in the relevant range, which on expanding the left-hand side using the double angle formula, and the right-hand side using the formula for $\sin (a+b)$ gives $$\left(\sin \frac x2-\sin \frac y2\right)\left(\cos \frac x2-\cos \frac y2\right)\le 0$$

Check that applies on the interval you need (the trigonometric functions of the half angles are positive). Then use this twice to get the result you are looking for.

$\endgroup$
1
$\begingroup$

You can easily show the fact that $\sin w$ is a concave function if you know calculus, because its second derivative on your interval is always less than or equal to $0$, (the second derivative is $-\sin w$)

$\endgroup$
  • 1
    $\begingroup$ It is concave on [0,π]. $\endgroup$ – Bernard Sep 27 '15 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.