# Prove using Jensen's Inequality

Let $\alpha_1, \alpha_2, . . . , \alpha_n$ be the interior angles of a convex (but not necessarily regular) n-gon. Prove, that for all integers $n\geq3$:

$$\cos \alpha_1 + \cos \alpha_2 + \cdots + \cos \alpha_n + n \cos\left(\dfrac{2\pi}{n}\right) \leq0$$

The prof said that I need to use Jensen's Inequality, and may be something else.. But I don't see it!

• Perhaps the last $+$ should be a $-$ ? Nov 12, 2013 at 18:12
• @leonbloy All are $+$ noting negative!
– user99638
Nov 12, 2013 at 18:14
• @leonbloy Doesn't help, for a quadrilateral that term is zero. Nov 12, 2013 at 18:15
• @Macavity With one angle larger than $\pi$ the polygon is not convex.
– Dirk
Nov 12, 2013 at 18:17
• There seems to be also a missing factor inside the last cosine. Check it Nov 12, 2013 at 18:19

If all the interior angles are acute, then noting that $\cos \theta$ is concave for $\theta \in (0, {\pi \over 2})$, we can use Jensen's Inequality.

$$\frac1n \sum \cos \alpha_i \le \cos \left(\frac1n \sum \alpha_i \right) = \cos \left( \frac{\pi(n-2)}{n} \right) = \cos(\pi - \frac{2\pi}{n}) = -\cos(\frac{2\pi}n)$$

Note that this is more restrictive than just being convex, and in fact your inequality does not hold in general if one of the angles can be larger than $\pi \over 2$.

• this Jensen's inequality is weird (different, let's say!)
– user99638
Nov 12, 2013 at 18:40
• What happened to the last term?
– user99638
Nov 12, 2013 at 18:42
• Well, I quite like it, it's useful in many cases especially when you have concave or convex functions... Nov 12, 2013 at 18:42
• Check, the last term is the RHS, in negative form. Nov 12, 2013 at 18:43
• ohh... I see... Definitely I got to read more an understand better Jensen's. It hard to apply it out of the blue without knowing the basis. THX!
– user99638
Nov 12, 2013 at 18:45