Let: $x,y,z >0$ and $ x+y+z=\dfrac{\pi}{2}$ then prove: $$\sin{x}\sqrt{1-\sin{x}}+\sin{y}\sqrt{1-\sin{y}}+\sin{z}\sqrt{1-\sin{z}} \ge 4\sqrt{2} \sin{x}\sin{y}\sin{z}(\sin{x}+\sin{y}+\sin{z})$$

I had a solution with Lagrange multipliers from this problem. But it is ugly and I'm thinking of a method without Lagrange multipliers but failed. I hope to see a nice solution.


Note that $$\sqrt{1-\sin x} = \sqrt{2}\sin\left(\frac{\pi}{4}-\frac{x}{2}\right) = \sqrt{2}\sin\left(\frac{y+z}{2}\right) \geq \frac{1}{2}\sqrt{2}\left( \sin y + \sin z\right).$$ Therefore $$\sin x \sqrt{1-\sin x}+\sin y \sqrt{1-\sin y} + \sin z \sqrt{1-\sin z} \geq \sqrt{2}\left(\sin x \sin y+ \sin x \sin z + \sin y \sin z\right) \\ = \sqrt{2}\sin x \sin y \sin z\left(\frac{1}{\sin x} + \frac{1}{\sin y} + \frac{1}{\sin z} \right) \\ \geq \sqrt{2}\sin x \sin y \sin z \frac{3}{\sin\frac{\pi}{6}}=6 \sqrt{2}\sin x \sin y \sin z$$ where the last inequality follows from the convexity of $\frac{1}{\sin}$. By concavity of $\sin$ $$\sin x + \sin y + \sin z \leq 3 \sin \frac{\pi}{6} = \frac{3}{2} $$ and your inequality follows.

  • $\begingroup$ very nice! I don't know how to deal with $\sin{\dfrac{y+z}{2}}$ before. now I find a more strong inequality. I am waiting more days to see if there is more surprising solutions. $\endgroup$ – chenbai Oct 20 '13 at 4:12

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