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- Feb 14, 2012

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Given $ A+B+C=\pi$, I need to prove $ cosA+cosB+cosC\leq \frac{3}{2}$.

I wish to ask if my following reasoning is correct.

First, I think of the case where A and B are acute angles, then I can use the Jensen's Inequality to show that the following is true.

$ cos\frac{A+B}{2}\geq \frac{cosA+cosB}{2}$

Carrying on with the working, I get

$ sin\frac{C}{2}\geq \frac{cosA+cosB}{2}$

$ 2sin\frac{C}{2}\geq cosA+cosB$

$ cosA+cosB\leq 2sin\frac{C}{2}$

$ cosA+cosB+cosC\leq 2sin\frac{C}{2}+cosC$

$ cosA+cosB+cosC\leq 2sin\frac{C}{2}+1-2sin^2C$

Completing square the RHS to obtain

$ cosA+cosB+cosC\leq -2(sin\frac{C}{2}-\frac{1}{2})^2+\frac{3}{2}$

Now, it's obvious to see that $ cosA+cosB+cosC\leq \frac{3}{2}$

My question is, can I solve the question by thinking A and B are acute angles and ignore the angle C right from the start so that I can let f(x)=cosx and notice that the curve of f(x)=cos x in the interval $x\in (0,\frac{\pi}{2})$ take the convex shape which in turn I can apply the Jensen's inequality without a problem?

Thanks.