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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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2 Answers 2

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
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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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