How to prove $6abc \lt a^2(b+c)+b^2(c+a)+c^2(a+b)$

Question

I am stuck on the following problem.

If $a,b,c >0$ and not all equal then prove that $$6abc \lt a^2(b+c)+b^2(c+a)+c^2(a+b)\lt 2(a^3+b^3+c^3). $$ Additional info:I'm looking for solutions using AM-GM .

My try:

$$a^2+b^2-ab \gt ab \implies a^3+b^3 \gt ab(a+b)=a^2b+ab^2$$. Similarly, $$b^2+c^2-bc \gt bc \implies b^3+c^3 \gt bc(b+c)=b^2c+bc^2$$ and $$c^2+a^2-ca \gt ca \implies c^3+a^3 \gt ca(c+a)=c^2a+ca^2$$.

By addition we get, $a^2(b+c)+b^2(c+a)+c^2(a+b)\lt 2(a^3+b^3+c^3)$

But I am stuck with

$6abc \lt a^2(b+c)+b^2(c+a)+c^2(a+b)$ part.

I will be grateful if someone explains . Thanks in advance ..

Answer 1

To solve the second inequality just expand and use AM-GM again \begin{equation}\sum_{cyc}a^2(b+c)=(a^2b+b^2c+c^2a)+(a^2c+c^2b+b^2a)\geq 3abc+3abc=6abc. \end{equation}

Answer 2

$a(b-c)^2+b(c-a)^2+c(a-b)^2>0$ gives the first inequality. And $(a+b)(a-b)^2+(b+c)(b-c)^2+(c+a)(c-a)^2>0$ gives the second inequality.

EDIT: To do it using AM-GM ... \begin{eqnarray*} a^2b+b^2c+c^2a> 3abc \\ a^3+a^3+b^3>3a^2b. \end{eqnarray*} & sum the appropriate permutations of these.

Answer 3

The right inequality.

By AM-GM for positives $x$ and $y$ we obtain: $$2x^3+y^3\geq3\sqrt[3]{x^6y^3}=3x^2y.$$ Thus, $$\sum_{cyc}2a^3=\frac{1}{3}\sum_{cyc}(2a^3+b^3+2b^3+a^3)\geq\frac{1}{3}\sum_{cyc}(3a^2b+3b^2a)=\sum_{cyc}(a^2b+ab^2)$$ and we are done!

The left inequality.

By AM-GM $$\sum_{cyc}(a^2b+a^2c)\geq6\sqrt[6]{\prod_{cyc}(a^2b\cdot a^2c)}=6abc$$ and we are done!