question from Euclid 2011 about proving that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$

Question

I just did the following question:

If $a, b, c$ positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ prove that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$

I solved it in the following way:

$ab+bc+ac=1$

From Tchebychev we have that $2(a^3+b^3)\ge (a^2+b^2)(a+b)$

So $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(a^3+c^3)b}{c^2+a^2}\ge \frac{(a+b)c+(b+c)a+(a+c)b}{2}$

$=ab+bc+ac=1$

This question really troubled me, as I had difficulties thinking of braking $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ into $ab+bc+ac=1$. Can someone please show me either a more intuitive approach to the question, or why I should have thought of braking up the original equation, into the second one, earlier and more intuitively?

Answer 1

SOS helps: $$\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-1=\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-ab-ac-bc=$$ $$=\sum_{cyc}\left(\frac{(a^3+b^3)c}{a^2+b^2}-\frac{c(a+b)}{2}\right)=$$ $$=\sum_{cyc}\frac{c(a+b)}{2}\left(\frac{2(a^2-ab+b^2)}{a^2+b^2}-1\right)=\sum_{cyc}\frac{c(a+b)(a-b)^2}{2(a^2+b^2)}\geq0.$$