0

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.

Community
  • 1
Apoorv Jain
  • 720
  • Rearrange the terms and you get the Schur's Inequality for $r=1$:

    $$a^3 + b^3 + c^3 + 3abc \ge a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$$

    The general form of Schur's Inequality is:

    $$a^r(a-c)(a-b) + b^r(b-a)(b-c) + c^r(c-a)(c-b) \ge 0$$

     – Stefan4024 Jun 15 '14 at 20:17

1 Answers1

2

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.

Anurag A
  • 41,067