How to prove $\frac{\sqrt{ab+bc+ca}}{2}\ge\frac{a^3(b+c)+b^3(c+a)+c^3(a+b)}{(a^2+b^2+c^2)(a+b+c)-2abc}.$?

Question

If $a,b,c\ge 0: a+b+c,>0$ then prove $$\frac{\sqrt{ab+bc+ca}}{2}\ge\frac{a^3(b+c)+b^3(c+a)+c^3(a+b)}{(a^2+b^2+c^2)(a+b+c)-2abc}.$$

I'm looking for a simple proof which student could full it in during test time.

I've tried to show it in SOS form and here is what I got \begin{align*} f(a,b,c)&=(ab+bc+ca)\left(\sum_{cyc}a^2.\sum_{cyc}a-2abc\right)^2-4\left(a^3(b+c)+b^3(c+a)+c^3(a+b)\right)^2\\ &=(ab+bc+ca)\prod_{\mathrm{cyc}}{\left( b+c-a \right) ^2}+4\sum_{\mathrm{cyc}}{b^3c^3\left( b-c \right) ^2}\ge 0. \end{align*} It is not nice as I expected. Can we make it more optimal ?

All idea is welcome.

Answer 1

Proof.

By multiplying $\sqrt{ab+bc+ca}$ for the OP, it is sufficient to prove$$(ab+bc+ca)\left[(a^2+b^2+c^2)(a+b+c)-2abc\right]\ge 2\sqrt{ab+bc+ca}\sum_{cyc}a^3(b+c) \tag{*}.$$Apply AM-GM as$$2\sqrt{ab+bc+ca}\sum_{cyc}(b+c)a^3\le \sum_{cyc}a^3\left[(b+c)^2+ab+bc+ca\right] \tag{1}.$$The result follows... Can you end it now ?

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Indeed, notice that$$\sum_{cyc}a^3(b+c)^2=(ab+bc+ca)\left(ab(a+b)+bc(b+c)+ca(c+a)-2abc\right),$$which implies\begin{align*} &\sum_{cyc}a^3\left((b+c)^2+ab+bc+ca)\right)\\ &=(ab+bc+ca)\left(a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)-2abc\right)\\ &=(ab+bc+ca)\left((a^2+b^2+c^2)(a+b+c)-2abc\right) \tag{2}. \end{align*} From $(1)$ and $(2)$ we get desired $(*).$

The proof is done. Equality holds at $a=b=t>0;c=0$ or $abc$ and cyclic permutations.