Proving Inequality for Variables with Constraints

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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: