<s>[INST] Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
I attempted to solve this question as follows:
$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le 3abc$
$(a+b)(b+c)(c+a)-2abc-3abc\le 3abc$
$(a+b)(b+c)(c+a)-5abc\le 3abc$
$(a+b)(b+c)(c+a)\le 8abc$
And I got stuck here. Could you please explain to me how to finish it off?
 [/INST] It is just Schur’s $^{[1]}$ Inequality of degree 3. You can search the web and read about it. It is pretty well known and someone named Aritra12 posted a handout on it recently on Aops.
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