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

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Discussion Overview

The discussion revolves around proving the inequality involving variables a, b, and c, constrained within the range of 0 to 1. The focus is on the mathematical reasoning required to establish the validity of the inequality.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Post 1 presents the inequality to be proven: $\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}$ for $a, b, c \in [0;1]$.
  • Post 2 reiterates the inequality and specifies that the domain is (0, 1), noting that the inequality fails if $a = b = c = 0$ or $a = b = c = 1$.
  • Post 3 also emphasizes the domain and the failure of the inequality under certain conditions, echoing Dan's earlier point.
  • Post 4 reveals a correction from Dan, indicating a previous misunderstanding regarding the direction of the inequality.
  • Post 5 offers a hint, suggesting that further exploration or insight may be forthcoming.
  • Post 6 indicates a proposed solution, though the details are not provided in the excerpt.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the inequality under specific conditions, with Dan acknowledging a mistake in his earlier assertion about the inequality's direction.

Contextual Notes

There are limitations regarding the assumptions made about the values of a, b, and c, particularly in relation to the endpoints of the interval [0, 1]. The discussion does not resolve the mathematical steps necessary to prove the 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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