You are given the sides of a triangle $a,b,c$ Prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ (Sweden 1950)

Question

You are given the sides of a triangle $a,b,c$ Prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

I was just doing the question above and solved it in the following manner:

$a^2b+a^2c-a^3+b^2c+b^2a-b^3+c^2a+c^2b-c^3-3abc\le 0$

$(a-b)(b-c)(c-a)+2b^2c+2ac^2+2a^2b-a^3-b^3-c^3-3abc\le0$

And this is where I got stuck. Could you please explain to me how to finish it off from here? Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$ I realise that this question has been asked before, but my main query, is how my train of thought could be used to finish off the question

Answer 1

WLOG $a\ge b\ge c$ this means $$c(a+b-c)\ge b(a+c-b)\ge a(c+b-a)$$

Now by rearrangement inequality $$\tag{1}a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le \sum_{cyc} ba(b+c-a)$$

Similarly, $$\tag{2}a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le \sum_{cyc} ca(b+c-a)$$

Adding $(1)$ and $(2)$ we complete our proof.

Answer 2

It's a triangle, so let \begin{eqnarray*} u&=&b+c-a \\ v&=&c+a-b \\ w&=&a+b-c \end{eqnarray*} and these quantities are non negative.

So your inequality is equivalent to \begin{eqnarray*} 2 (u(v+w)^2+v(w+u)^2+w(u+v)^2) \leq 3(u+v)(v+w)(w+u) \end{eqnarray*} or \begin{eqnarray*} 2 \sum u^2 v +12 uvw \leq 3 \sum u^2 v +6 uvw \end{eqnarray*} which is obvious.

Answer 3

Because your way does not give a solution.

We can use SOS here.

We need to prove that: $$\sum_{cyc}(a^2b+a^2c-a^3)\leq3abc$$ or $$\sum_{cyc}(a^2b+ab^2-a^3-b^3)\leq\sum_{cyc}(abc-a^3)$$ or $$\sum_{cyc}(a^3-abc)\leq\sum_{cyc}(a-b)^2(a+b)$$ or $$(a+b+c)\sum_{cyc}(a^2-ab)\leq\sum_{cyc}(a-b)^2(a+b)$$ or $$(a+b+c)\sum_{cyc}(a-b)^2\leq2\sum_{cyc}(a-b)^2(a+b)$$ or $$\sum_{cyc}(a-b)^2(a+b-c)\geq0$$ and we are done!