$a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right) \geq 3\left(a^3b+b^3c+c^3a\right)$

Question

Let: $a,b,c >0$. Prove that:

$$a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right) \geq 3\left(a^3b+b^3c+c^3a\right)$$

I read an ugly solution:

Take $LHS-RHS$, get:

$$\sum\left(-ab+ac-5bc\right)\left(a-b\right)^2 \geq 0$$

It is by Maple, but how can a person solve this? Thanks

Answer 1

The inequality is wrong.

Try $a=b=1$ and $c\rightarrow0^+$.

The reversed inequality is true for any triangle.

Indeed, we need to prove that: $$3(a^3b+b^3c+c^3a)\geq a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c),$$ which is true because $$\sum_{cyc}(3a^3b-a^2b^2-2a^2bc)=\frac{1}{2}\sum_{cyc}(6a^3b-2a^2b^2-4a^2bc)=$$ $$=\frac{1}{2}\sum_{cyc}(3a^3b+3a^3c-6a^2b^2+4a^2b^2-4a^2bc+3a^3b-3a^3c)=$$ $$=\frac{1}{2}\sum_{cyc}(3ab(a-b)^2+2c^2(a-b)^2+3(a+b+c)(a^2b-a^2c))=$$ $$=\frac{1}{2}((a-b)^2(3ab+2c^2)-(a+b+c)(a-b)^3)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(3ab+2c^2-a^2+b^2-ac+bc)\geq$$ $$\geq\frac{1}{2}\sum_{cyc}(a-b)^2(c^2-a^2+b^2)$$ and it's enough to prove that: $$\sum_{cyc}(c^2-a^2+b^2)\geq0,$$ which is obvious and $$\sum_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)\geq0,$$ which is true because $$\sum_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)=\sum_{cyc}(2a^2b^2-a^4)=$$ $$=(a+b+c)\prod_{cyc}(a+b-c)>0.$$

Answer 2

As others have pointed out, the inequality is wrong in the given direction. However, it is true in the reverse direction.

Let $f(a,b,c) = \sum_{cyc} 3a^3b -a^2b^2- 2a^2bc$, and assume $a$ is the largest wlog, since $f$ is cyclic. If $c > b$ then $$ f(a,c,b) - f(a,b,c)= 3(a^3c - a^3b + c^3b-b^3c+b^3a-c^3a)=3(a+b+c)(a-b)(a-c)(c-b)\geq 0, $$ thus we can assume wlog that $a\geq b\geq c$. By Muirhead we have $$ 2\sum_{cyc}a^2bc\leq\sum_{sym}a^3b\leq 2\sum_{cyc}a^3b, $$ and $$ \sum_{cyc}a^2b^2=\frac{1}{2}\sum_{sym}a^2b^2\leq \frac{1}{2}\sum_{sym}a^3b\leq\sum_{cyc}a^3b, $$ thus $f(a,b,c)\geq 0$.