How do I prove the following inequality?

Question

How do I proceed to solve the inequality
$$\frac{(a^2+b^2)}{(a+b)} + \frac {(b^2+c^2)}{(b+c)} + \frac{(a^2+c^2)}{(a+c)} \geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $\frac {(a^2+b^2)}{(a+b)}$ to $AM$.
I have tried finding $a_1$ and $a_2$ by doing $ (\frac{(a_1+a_2)}{2}\bigr)^2 \geq$ $\frac{(a^2+b^2)}{(a+b)}$

Answer 1

If $a$, $b$, $c>0$ then $$\frac{a^2+b^2}{a+b}\ge\frac{a+b}2$$ etc.

Answer 2

Using Cauchy Schwarz in Engel form, we have

$$\!\!\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}+ \frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{a+c}\geq \frac{(a+b+c)^2}{2(a+b+c)}+\frac{(b+c+a)^2}{2(a+b+c)}= a+b+c$$