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<a and b^2<b (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


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