# How to solve this problem about the reciprocals of the angles inside an acute triangle?

I was solving some previous year papers and stumbled upon this question. I don't even know how to approach this problem. The problem is as follows:

Consider an acute angled triangle $PQR$ such that $C$, $I$ and $O$ are the circumcentre, incentre and orthocentre respectively. Suppose $\angle QCR$, $\angle QIR$ and $\angle QOR$, measured in degrees, are $\alpha$, $\beta$ and $\gamma$ respectively. Show that $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}>\frac{1}{45^\circ}$$

Sorry, but I am unable to show any progress as I haven't made any. There aren't any trigonometric functions and I cannot find a way to handle the raw angles. At some point, it seemed that polynomials might work. But, that'll be a bit too weird. So, it'll be great if someone can help.

• reciprocal of an angles? – Narasimham Apr 5 '16 at 4:27
• @Narasimham Yes. – SinTan1729 Apr 5 '16 at 4:31

Since no one else answered it, and I've managed to solve it, I decided to post an answer myself. Consider the following graph: Clearly, $\alpha=2P$ and $\beta=\frac{2P+Q+R}{2}$ and $\gamma=Q+R$.
Now, $\frac{1}{x}$ is a convex function. So, from Jensen's inequality, we get $$\frac{\frac{1}{2P}+\frac{1}{2P+Q+R}}{2}>\frac{2}{\frac{1}{Q+R}}$$
So, $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{1}{2P}+\frac{1}{2P+Q+R}+\frac{1}{Q+R}>\frac{6}{2P+Q+R}=\frac{6}{180^\circ+P}>\frac{6}{270^\circ}=\frac{1}{45^\circ}$