# maximum value $P= \sqrt{2}\sin \frac{A}{2}+ \sqrt{3}\sin \frac{B}{2} + \sqrt{6}\sin \frac{C}{2} + \sqrt{3}\sin \frac{D}{2}$

Let $ABCD$ be a convex quadrilateral. what is the maximum value: $$P= \sqrt{2}\sin \frac{A}{2}+ \sqrt{3}\sin \frac{B}{2} + \sqrt{6}\sin \frac{C}{2} + \sqrt{3}\sin \frac{D}{2}$$

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. – Julian Kuelshammer Oct 24 '12 at 6:36

Hint

Use the fact that $|\sin(x)|\leq 1\,.$

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What does this mean? – Cameron Buie Oct 24 '12 at 7:04
@CameronBuie:The post is under the tag "inequality". – Mhenni Benghorbal Oct 24 '12 at 8:00
Yes...but how is simply knowing that each of $\sin\frac{A}2,...,\sin\frac{D}2\leq\frac1{\sqrt{2}}$ sufficient to maximize $P$? – Cameron Buie Oct 24 '12 at 9:27

There are four unknown angles $\alpha, \beta,\gamma,\delta\in\bigl[0,{\pi\over2}\bigr]$ that satisfy the constraint $\alpha+ \beta+\gamma+\delta=\pi$. Looking at the expression to maximize we note that $$\sqrt{3}(\sin\beta+\sin\delta)\leq 2\sqrt{3}\sin{\beta+\delta\over2}\ ,$$ which shows that at the max we necessarily have $$\beta=\delta\qquad\left(={\pi\over2}-{\alpha+\gamma\over2}\right)\ .$$ Now put $$\alpha:=\mu-\lambda\ ,\quad \gamma:=\mu+\lambda\ .$$ Then we have to maximize the function $$g(\mu,\lambda):=(\sqrt{6}+\sqrt{2})\sin\mu\cos\lambda +(\sqrt{6}-\sqrt{2})\cos\mu\sin\lambda+2\sqrt{3}\cos\mu\ .$$ Now a function $$p(\lambda):=a\cos\lambda+ b\sin\lambda$$ has maximal value $A:=\sqrt{a^2+b^2}$, and this value is taken at the points $\lambda=\arg(a,b)$. This observation allows us to eliminate $\lambda$ from $g$, so that we now have to maximize $$q(\mu):=\sqrt{(\sqrt{6}+\sqrt{2})^2\sin^2\mu +(\sqrt{6}-\sqrt{2})^2\cos^2\mu} +2\sqrt{3}\cos\mu$$ by proper choice of $\mu$. Putting $\cos\mu=:z$ we obtain after some calculation $$\tilde q(z)=\sqrt{8+4\sqrt{3}-8\sqrt{3}z^2}+2\sqrt{3} z\ ,$$ which, low and behold, takes its max at $z={1\over2}$! I may leave the rest of the computation to you.

After all the necessary calculations have been completed it remains to check whether we have indeed found the global max, in particular, whether to the found $\mu$ belongs an admissible $\lambda$.

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