Proving Inequality for Variables with Constraints

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    Inequality Proof
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Discussion Overview

The discussion centers around proving an inequality involving three variables \(a\), \(b\), and \(c\) constrained between 0 and 1. Participants are exploring the validity of two different forms of the inequality and providing solutions or corrections related to it.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant proposes the inequality: \(\sqrt{a(1-b)(1-c)} + \sqrt{b(1-a)(1-c)} + \sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}\).
  • Another participant suggests a different form of the inequality: \(\sqrt{a(1-b)(1-c)} + \sqrt{b(1-a)(1-c)} + \sqrt{c(1-a)(1-b)} \ge 1 + \sqrt{abc}\).
  • A participant acknowledges a typo in their earlier post, indicating that the inequality sign should be reversed, but does not clarify which form is correct.
  • One participant expresses appreciation for another's solution, indicating a positive reception to the contributions made.

Areas of Agreement / Disagreement

Participants do not reach consensus on the correct form of the inequality, as one proposes an upper bound while another suggests a lower bound. The discussion remains unresolved regarding which inequality is valid.

Contextual Notes

There is a noted typo in one participant's post that affects the interpretation of the inequality, but the specific implications of this correction are not fully explored.

lfdahl
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Let $0 \le a,b,c \le 1.$ Prove the inequality:$\sqrt{a(1-b)(1-c)}+ \sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}$
 
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lfdahl said:
Let $0 \le a,b,c \le 1.$ Prove the inequality:$\sqrt{a(1-b)(1-c)}+ \sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \ge 1 + \sqrt{abc}(1)$
for a=b=c=1 ,(1) is not true
 
Albert said:
for a=b=c=1 ,(1) is not true

You´re right, Albert. I´ve made a typo. The inequality sign should be reversed. I´m sorry for my mistake.

Thankyou for pointing this out to me.

Cheers, lfdahl
 
lfdahl said:
Let $0 \le a,b,c \le 1.$ Prove the inequality:$\sqrt{a(1-b)(1-c)}+ \sqrt{b(1-a)(1-c)}+\sqrt{c(1-a)(1-b)} \le 1 + \sqrt{abc}$
my solution:
$set:\,\, (1-a)=sin^2 A,(1-b)=sin^2 B, 1-c=sin^2C,it\,\,is\,\,to\,\,prove:\\
cosAsinBsinC+cosBsinCsinA+cosCsinAsinB-cosAcosBcosC\leq1\\
or\,\, cosA(sinBsinC-cosBcosC)+sinA(sinBcosC+cosBsinC)\leq 1\\
sinAsin(B+C)-cosAcos(B+C)\leq 1\\
cosAcos(B+C)-sinAsin(B+C)\geq -1\\
cos(A+B+C)\geq -1$
$and \,\, the\,\,proof\,\,is \,\, done $
 
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Albert said:
my solution:
$set:\,\, (1-a)=sin^2 A,(1-b)=sin^2 B, 1-c=sin^2C,it\,\,is\,\,to\,\,prove:\\
cosAsinBsinC+cosBsinCsinA+cosCsinAsinB-cosAcosBcosC\leq1\\
or\,\, cosA(sinBsinC-cosBcosC)+sinA(sinBcosC+cosBsinC)\leq 1\\
sinAsin(B+C)-cosAcos(B+C)\leq 1\\
cosAcos(B+C)-sinAsin(B+C)\geq -1\\
cos(A+B+C)\geq -1$
$and \,\, the\,\,proof\,\,is \,\, done $

Very nice solution, Albert! :cool:
 

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