Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
 A: By AM-GM inequality,
$$\frac{a^2}{a+b} + \frac{a+b}{4} \ge a$$
Add up the similar inequalities obtained by cyclic substitution, you are done.
A: This is just Cauchy-Schwarz:
$$\left(\frac{a}{\sqrt{a+b}}\sqrt{a+b}+\frac{b}{\sqrt{b+c}}\sqrt{b+c}+\frac{c}{\sqrt{a+c}}\sqrt{a+c}\right)^2 \leq \left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} \right)\big( (a+b)+(a+c)+(b+c)\big)$$
A: Cauchy-Swartz:
$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+ \cdots +\frac{a_n^2}{b_n} \geq \frac{(a_1+a_2+ \cdots +a_n)^2}{b_1+b_2+ \cdot +b_n}$
So,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$
A: Hint: $ \sum \frac{a^2 - b^2}{a+b} = \sum (a-b) = 0$.
(How is this used?)
Hint: $\sum \frac{a^2 + b^2}{a+b} \geq \sum \frac{a+b}{2} = a+b+c$ by AM-GM. 
Hence, $\sum \frac{ a^2}{ a+b} \geq \frac{1}{2}(a+b+c)$. 
A: The Cauchy-Schwarz inequality implies that
$$\left(\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a}\right)\left((a+b)+(b+c)+(c+a)\right)\geq(a+b+c)^2,$$
i.e.
$$\left(\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a}\right)\times 2\geq a+b+c,$$
which gives the result.
A: You can write $\frac{a^2}{a+b} = a - \frac{ab}{a+b}$, and similarly with the other terms.
It remains to prove (after rearranging the inequality) that:
$$\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leq \frac{1}{2}\left(a+b+c\right)$$
which follows from AM-HM.
