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
Best Answer
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 $$\sum_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)=\sum_{cyc}(2a^2b^2-a^4)=$$ $$=(a+b+c)\prod_{cyc}(a+b-c)>0.$$