TITLE: Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$
QUESTION [6 upvotes]: Let $a_1, a_2, a_3, \dots , a_n$ be positive real numbers whose sum is $1$. Prove that
$$\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \ldots +\frac{a_n^2}{a_n+a_1} \geq \frac12\,.$$

I thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $2$ (the denominator on the right hand side). I would really appreciate any hints.

REPLY [6 votes]: The following modified form of the CBS inequality is often helpful:

Lemma. If $x_i \in \Bbb{R}$ and $a_i > 0$, then
  $$ \frac{x_{1}^{2}}{a_{1}} + \cdots + \frac{x_{n}^{2}}{a_{n}} \geq \frac{(x_{1} + \cdots + x_{n})^{2}}{a_{1}+\cdots+a_{n}}. $$

The proof is straightforward using the CBS inequality, following the methodology exactly John Marty used.
Applying this, we have
$$ \frac{a_{1}^2}{a_{1}+a_{2}} + \cdots + \frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{(a_{1} + \cdots + a_{n})^{2}}{2(a_{1}+\cdots+a_{n})} = \frac{1}{2}. $$