MHB Proving Inequality for Variables with Constraints

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    Inequality Proof
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The discussion revolves around proving two inequalities involving variables a, b, and c constrained between 0 and 1. The first inequality states that the sum of square roots of products involving a, b, and c should be less than or equal to 1 plus the square root of their product. A typo was acknowledged regarding the direction of the inequality, which was later corrected. Participants expressed appreciation for the solutions provided, highlighting the collaborative nature of the discussion. Overall, the focus remains on establishing the validity of these mathematical inequalities.
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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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