## Cute inequality

Here is a problem that appeared in Romanian Mathematical Olympiad (Junior Team Selection Test 2002).

If $a,b,c\in (0,1)$ prove that

$\displaystyle \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$

First Solution: Use Cauchy-Schwarz inequality to obtain:

$\displaystyle \left(\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\right)^2 \le \left[ab+(1-a)(1-b)\right]\left(c+(1-c)\right)=$

$\displaystyle =2ab-a-b+1\le a^2+b^2-a-b+1=(a^2-a)+(b^2-b)+1<1$

Here we used the fact that $2ab\le a^2+b^2$ (which follows from $(a-b)^2\ge 0$), $a^2 and $b^2 (which follows from $a,b\in (0,1)$). Taking square root of both sides, we obtain the desired inequality.

Second Solution: Since $a,b,c\in (0,1)$, there exists $x,y,z\in (0,\pi/2)$ such that

$\displaystyle a=\sin^2(x) \ \ \ \ b=\sin^2(y) \ \ \ \ c=\sin^2(z)$

Thus, the problem now reads as

$\displaystyle \sin(x)\sin(y)\sin(z)+\sqrt{(1-\sin^2(x))(1-\sin^2(y))(1-\sin^2(z))} < 1$

or equivalently,

$\displaystyle \sin(x)\sin(y)\sin(z)+\cos(x)\cos(y)\cos(z) < 1$

We can prove this inequality as follows

$\displaystyle \sin(x)\sin(y)\sin(z)+\cos(x)\cos(y)\cos(z) <\sin(x)\sin(y)+\cos(x)\cos(y)=\cos(x-y)\le 1$

Done!