<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] $a=y+z,b=x+z,c=x+y$ then we have to prove $$\sum 2z{(y+z)}^2\le 3(x+y)(y+z)(z+x) \tag 1$$ Now note that $$\sum (2x{(z+y)}^2-8xyz)=\sum 2x{(y-z)}^2$$ and $$(x+y)(y+z)(x+z)-8zxy=\sum z{(x-y)}^2$$ therfore  after subtracting $24xyz$ from both sides of $(1)$ it remains to rpove $$\sum z{(x-y)}^2\ge 0$$ which is true
Note that one could actually multiply out inequality $(1)$ to get $$x^2(y+z)+y^2(z+x)+z^2(x+y)\ge 6xyz$$ which is true by AM_GM but i felt writing it as an SOS is more nice
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