STEP II 2007 question 7 solution

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Since

, and in , is concave in this interval.

Similarly,

,

which is negative for , (and indeed negative for as well, the only problem being is not defined there), and thus is concave in .

(i) First we note that since they are angles in a triangle.

Applying Jensen's inequality with , and yields

.

But since , the RHS equals , and thus we have

from which the desired result immediately follows.

(Note that since are angles in a triangle, and so our usage of Jensen's inequality was valid.)

(ii)

Setting, in Jensen's inequality, and , we get

Exponentiating both sides (note that since is an increasing function, this preserves the inequality)

as required.

a) Setting, in , we get

from which follows (note that we in this step use

as required.

(Again, the usage of Jensen's inequality was valid, since which was a requirement for the function to be concave)

b) Let .

Set, in [b](*)[/b], , , and we get

(This inequality is valid whenever are positive, because then are positive as well, which was a requirement for [b](*)[/b] to be valid)

Now, this means

.

But attains this value at , which means that this is the function's minimum. (By definition, since we have just proven that for all , where , we have )

and so the desired minimum value for the expression is -2.

Solution by ukgea.