Jensen inequality conceptual doubt

Question

Prove that in a triangle $ABC$,$\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}\geq\frac{3}{4}$.

I tried to solve it by Jensen's inequality.I let $f(x)=\sin^2\frac{x}{2}$

$f''(x)=\frac{1}{2}\cos x>0$ in $(0,\frac{\pi}{2})$,so it is a convex function and the given inequality can be proved easily by Jensen's inequality.But my doubt is as the graph of $\sin^2\frac{x}{2}$ is a concave function(downward pointing cups) whereas its second derivative tells that it is a convex function.Why is this ambiguity exist?

http://www.wolframalpha.com/input/?i=plot+%28sin%28x%2F2%29%29%5E2+from+0+to+pi%2F2 — 5xum, Sep 03 '15 at 11:45
Perhaps you are looking at the graph of $\sin$ and not the graph of $\sin^2$ ? — Sergio Parreiras, Sep 03 '15 at 11:57
@Vinod, you can try this method, too : http://math.stackexchange.com/questions/1419652/prove-that-in-triangle-abc-cos2a-cos2b-cos2c-geq-frac34/1419674#1419674 — lab bhattacharjee, Sep 03 '15 at 13:06

score 2 · Accepted Answer · edited Sep 03 '15 at 11:51

2

The function is convex, so the graph is an upward pointing cup (for example, $f(x)=x^2$ is also convex and is an upward pointing cup.

There is no ambiguity, $\sin^2\frac x2$ is convex. You may have doubts, but you can hardly argue with the proofs.

Also, you can "see" here that the function is convex.

edited Sep 03 '15 at 11:51

Michael Hardy

1

answered Sep 03 '15 at 11:48

5xum

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Jensen inequality conceptual doubt

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