If a, b, c > 0 and a + b + c = 3, prove that abc =< 1

leviethoang1012

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help me

Prove that if a, b, c is the length of three sides of a triangle have equal perimeter, then 3: 3a + 3b ^ 2 ^ 2 ^ 2 + 3c + 4abc ≥ 13
 
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You can find the maximum of abc given the constraints (a+b+c-3)=0, and a,b,c>0.
Use a Lagrange multiplier. Given the solution for the maximum of abc it's clear that abc<=1.

This is posted in "Beginning Algebra." But from the looks of the problem, it belongs in the "Intermediate/Advanced Algebra" section.
We would do well to try to speak to the level of the "audience" of the students asking for help.

I have used the inequality Cosi but do not make results

I think the wrong problem

leviethoang1012, you mentioned an inequality with some shortened form of a name, but I don't know what
you mean by it. What level math are you in specifically?
 
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