Inequality, a3+b3+c3+3abc ≥ a2b+a2c+b2a+b2c+c2a+c2b

Sesame

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a,b,c >0, proof that a^3+b^3+c^3+3abc ≥ a^2b+a^2c+b^2a+b^2c+c^2a+c^2b
 
Please post what you know with the problem. Please do not post the problem in the title. As it is, the exponentiation did not show up in the title, which defeats the partial purpose of the expectation in the title.
 
I was hoping (against hope) that OP will check that after expanding!!
My apologies. I failed to use mind-reading.
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Did you copy the question correctly?
Subhotosh Khan's hint will apply if the sign in front of 3abc is negative.
If we made the sign in front of 3abc negative then the inequality wouldn't be true.

A counterexample would be a=b=c=1
LHS a^3 + b^3 + c^3 - 3*a*b*c = 0
RHS a^2*b + a^2*c + b^2*a + b^2*c + c^2*a + c^2*b = 6
 
If we made the sign in front of 3abc negative then the inequality wouldn't be true.
Indeed. That's why I questioned whether SK's hint was applicable.
I managed to prove the inequality using a few facts starting with expanding this one.

\(\displaystyle (a+b+c)(a^2+b^2+c^2)\)
 
Indeed. That's why I questioned whether SK's hint was applicable.
I managed to prove the inequality using a few facts starting with expanding this one.

\(\displaystyle (a+b+c)(a^2+b^2+c^2)\)
This problem was solved many many full-moons ago, in this forum. Since the OP did not show any work, I was trying nudge OP to try factorization. Beyond that, the problem statement was ill-formatted and that needed to be fixed.
 
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