<s>[INST] Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$ 
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.
 [/INST] Thanks to Sanchez for giving me a hint to solve this. Here is a full solution.
By the Cauchy-Schwarz inequality we have:
$${\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1}=\frac{a_1^2}{(\sqrt{a_1+a_2})^2}+\frac{a_2^2}{(\sqrt{a_2+a_3})^2}+ \cdots+ \frac{a_n^2}{(\sqrt{a_n+a_1})^2} \geq \frac{1}{a_1+\cdots + a_n+a_1+ \cdots + a_n}\left(\frac{a_1 \cdot \sqrt{a_1+a_2}}{\sqrt{a_1+a_2}} + \frac{a_2 \cdot \sqrt{a_2+a_3}}{\sqrt{a_2+a_3}}+ \cdots + \frac{a_n \cdot \sqrt{a_n+a_1}}{\sqrt{a_n+a_1}}\right)\\=\frac{a_1+a_2+a_3+ \cdots a_n}{{2(a_1+a_2+a_3+ \cdots a_n)}}=\frac12}$$ 
as required. (We know that $a_1+a_2+a_3+ \cdots +a_n=1$) 
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