Let $a,b,c$ be positive numbers. Prove that $a^2b+b^2c+c^2a\leq a^3+b^3+c^3$

Question

I tried moving all the variables onto one side and attempting to factor it into squares and proving the inequality since squares are always greater than or equal to 0. The inequality is too complex to factor through completing the square (or "rectangle"). How do I solve the inequality?

Answer 1

Since $(a^2,b^2,c^2)$ and $(a,b,c)$ are the same orderd, it's just Rearrangement: $$a^3+b^3+c^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq a^2b+b^2c+c^2a$$ Also we have the following SOS: $$\sum_{cyc}(a^3-a^2b)=\sum_{cyc}\left(a^3-a^2b-\frac{1}{3}(a^3-b^3)\right)=\frac{1}{3}\sum_{cyc}(a-b)^2(2a+b)\geq0$$

Answer 2

Apply the Arithmetic Mean-Geometric Mean to $a^3,a^3,b^3$: $$\frac{a^3+a^3+b^3}3\geq\sqrt[3]{a^3a^3b^3}$$

Answer 3

An alternative solution. We may define $$g(a,b,c) = a^2(a-b)+b^2(b-c)+c^2(c-a) $$ and check that if we assume that $b,c$ are fixed, the minimum of the previous expression occurs at $a=\frac{1}{3}\left(b+\sqrt{b^2+3c^2}\right)$, by solving $\frac{\partial g}{\partial a}=0$ and noticing that $\frac{\partial g}{\partial a}(0)\leq 0$. If we assume that $a$ and $c$ are fixed, the minimum occurs at $b=\frac{1}{3}\left(c+\sqrt{c^2+3a^2}\right)$ and if we assume that $a,b$ are fixed the minimum occurs at $c=\frac{1}{3}\left(a+\sqrt{a^2+3b^2}\right)$. If all these constraints hold, $a\leq \max(b,c)$, $b\leq\max(a,c)$ and $c\leq\max(a,b)$, hence $a=b=c$. In such a case $g$ equals zero, hence in the general case $g\geq 0$ as wanted.