If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$

Question

$a$, $b$, $c$ are sides of a triangle, prove: $$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$

What I have tried: $$ ⇔2\sum (a+b)ab\geq \sum a^3+9abc $$ so I can't use $$\sum a^3+3abc\geq \sum ab(a+b)$$

Answer 1

WLOG $$c=\max\{a,b,c\}\implies a=x+u,b=x+v\text{ and }c=x+u+v (x>0, u,v\ge 0)$$

Then your inequality is equivalent to

$$x(u^2-uv+v^2)+2(u+v)(u-v)^2\ge 0*\text{true}*$$

Answer 2

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

Answer 3

Also, we can use SOS here.

Indeed, by your work we need to prove that $$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)\geq0.$$

Now, let $a\geq b\geq c$.

Thus, since $$3b-a-c=b+c-a+2b-2c>0,$$ we obtain: $$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)=\sum_{cyc}(2a^2b+2a^2c-4abc-a^3+abc)=$$ $$=\sum_{cyc}\left(2c(a-b)^2-\frac{1}{2}(a+b+c)(a-b)^2\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(3c-a-b)\geq$$ $$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-c)^2(3b-a-c))\geq$$ $$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-b)^2(3b-a-c))=(a-b)^2(b+c-a)\geq0.$$