Prove inequality in a triangle

Question

Prove that

In a triangle having sides $a, b, c$

$$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc.$$

I tried using the basic two side sum greater than third property but got nothing hope you guys help me.

Answer 1

The inequality given can be rewritten as $$ a^3+b^3+c^3+3abc \geq a^2(b+c)+b^2(c+a)+c^2(a+b).$$ Without the loss of generality assume that $a \geq b \geq c$. Now observe that \begin{align*} a^3+b^3+c^3+3abc -a^2(b+c)-b^2(c+a)-c^2(a+b) & = (a-b)^2(a+b-c)+c(a-c)(b-c). \end{align*} The expression on the right is always $\geq 0$ by the assumption that $a \geq b \geq c$. Hence the inequality holds.

PS: observe that the inequality is true without having $a,b,c$ to satisfy the triangle inequality.