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

AI Thread Summary
The discussion centers on proving the inequality $\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}$ for variables a, b, and c within the range of 0 to 1. It is noted that the inequality does not hold when a, b, and c are all 0 or all 1. A participant expresses doubt about the validity of the statement, suggesting that the inequality may actually go the other way. The conversation highlights the importance of careful consideration of boundary conditions in mathematical proofs. Ultimately, the focus remains on finding a valid proof for the stated inequality.
lfdahl
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Prove the inequality:

$\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}, \;\;\;\;a,b,c \in [0;1].$
 
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lfdahl said:
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
 
topsquark said:
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?
 
(Swearing) I had the inequality going the other way!

Thanks for the catch.

-Dan
 
Hint:

Use a trigonometric substitution
 
Here´s the suggested solution:

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.
 
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