I am currently studying convex functions and came across the following exercise.
Let $D = (-\frac{\pi}{2},0) \times (-\frac{\pi}{2},0) \subset \mathbb R^2$. Consider the function $f$ defined by $f(x,y) = \sin(x+y),$ for every $(x,y) \in D.$
I was able to show that $D$ is convex, $f$ is convex and also showed that every point of the form $(-(y+\frac{\pi}{2}),y)$ is a global minimizer of $f$, for $y \in (-\frac{\pi}{2},0).$
Exercise. Show that the inequality
$$ 3\sin\left(\frac{1}{3}(x+y)+\frac{2}{3}(z+w) \right) \leqslant \sin(x+y)+2\sin(z+w)$$
is valid, for every $-\pi/2 < x,y,z,w < 0.$
My attempt. The only idea that comes up to my mind while solving this exercise is to use the gradient inequality (since $f$ is convex). This inequality tells us that
$$ f(a)-f(b) \geqslant \nabla f(b)^\top(a-b), $$ for every elements $a,b \in D.$ Denoting $a = (x,y)$ and $b=(z,w)$, the inequality above yields
$$ \sin(x+y)-\sin(z+w) \geqslant \cos(z+w)( x+y - (z+w) ).$$ And I really don't see how this can be transformed into the inequality presented in the exercise, altought there are some obvious similarities: I have tried using the formula for $cos(z+w)$ but think this doesn't help either - so maybe this isn't the correct approach?
Thanks for any help in advance.
As you already showed that $f$ is convex set $a:=(x+y)$ and $b:=(z+w)$, Then you just have to prove $$ 3 f(\frac{1}{3}a+\frac{2}{3}b) \leq f(a) + 2f(b)$$ But due to the convexity of $f$ this is true.