Show that for every numbers a,b,c real positive we have $\sum a\frac{b^2+c^2}{b+c}\geq ab+bc+ca$

Question

Show that for every numbers $a,b,c$ real positive we have

$$\sum a\frac{b^2+c^2}{b+c}\geq ab+bc+ca$$

That $a$ in front is really annoying so I tried: $abc$ and we get that

$$\sum \frac{b^2+c^2}{bc(b+c)}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

I am pretty sure that we have to use the inequality of means and I tried writing some inequality from this, but got to nothing useful. Hope one of you can help me! Thank you!

Answer 1

Note that $$\frac{a^2+b^2}{a+b}\ge \frac{a+b}2,$$ since $$2(a^2+b^2)\ge(a+b)^2,$$ or $$a^2+b^2\ge2ab.$$

Hence $$\frac{a^2+b^2}{a+b}\ge \frac{a+b}2$$ $$\frac{b^2+c^2}{b+c}\ge \frac{b+c}2$$ $$\frac{c^2+a^2}{c+a}\ge \frac{c+a}2.$$

So we have $$c\cdot\frac{a^2+b^2}{a+b}\ge c\cdot\frac{a+b}2.$$ $$a\cdot\frac{b^2+c^2}{b+c}\ge a\cdot\frac{b+c}2$$ $$b\cdot\frac{c^2+a^2}{c+a}\ge b\cdot\frac{c+a}2.$$

Add these and get the desired inequality.

Answer 2

It is straightforward to verify that $$ \frac{x^2+y^2}{x+y} \ge \frac 12 (x+y) $$ holds for all positive real numbers $x, y$. This gives $$ \sum_{cyc} a \frac{b^2+c^2}{b+c} \ge \frac 12 \bigl( a(b+c)+b(c+a)+c(a+b) \bigr) = ab+bc+ca \, . $$