Prove $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$

Question

if $a,b,c$ are positive real numbers,Prove:$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$

Things I have done so far: I know the fact that $$a^3+b^3+c^3\geq\frac{1}{2}[ab(a+b)+bc(b+c)+ca(c+a)]$$

However i tried to reach the problem inequality,but I was not succesful.

Source of problem: school exam.

@Aditya,it was not school main exam,it was a exam for students who study for mathematical olympiad. — user2838619, Aug 17 '14 at 12:47
Alternately, try Rearrangement inequality with $(a^2, b^2, c^2)$ and $(a, b, c)$. — Macavity, Aug 17 '14 at 13:26

score 5 · Accepted Answer · answered Aug 17 '14 at 12:43

5

By aM-GM: $$\frac{2a^3+b^3}3\ge(a^6b^3)^{\frac13}=a^2b$$ $$\frac{2b^3+c^3}3\ge(b^6c^3)^{\frac13}=b^2c$$ $$\frac{2c^3+a^3}3\ge(c^6a^3)^{\frac13}=c^2a$$ Add them.

answered Aug 17 '14 at 12:43

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Prove $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$

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