How to Prove the Inequality for a, b, and c in the Range of 0 to 1?

[lfdahl]

Prove the inequality:

$\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}, \;\;\;\;a,b,c \in [0;1].$

 

[topsquark]

Prove the inequality:

$\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}, \;\;\;\;a,b,c \in [0;1].$

Your domain is (0, 1). If a = b = c = 0 or if a = b = c = 1 the inequality is false.

-Dan

 

[lfdahl]

Your domain is (0, 1). If a = b = c = 0 or if a = b = c = 1 the inequality is false.

-Dan

Hi, Dan!

Are you quite sure in your statement?

 

[topsquark]

(Swearing) I had the inequality going the other way!

Thanks for the catch.

-Dan

 

[lfdahl]

Hint:

Spoiler

Use a trigonometric substitution

 

[lfdahl]

Here´s the suggested solution:

Spoiler

Since $a,b,c \in [0;1]$, we have:
\[\exists x,y,z \in \left [ 0;\frac{\pi}{2} \right ]:\: \: \sin^2x = a, \: \: \sin^2y = b\: \: and\: \: \sin^2z = c.\]The inequality then reads:\[\sin x\cos y\cos z + \sin y \cos x\cos z + \sin z\cos x\cos y \leq 1 + \sin x\sin y\sin z\]\[ \cos z(\sin x \cos y + \sin y \cos x)+\sin z (\cos x \cos y - \sin x \sin y) \leq 1 \]
$\;\;\;\;\; \cos z \sin (x+y)+\sin z \cos (x+y) = \sin(x+y+z) \leq 1$, and we´re done.