Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$

Solution 1:

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 $a_1+a_2+a_3+ \cdots +a_n=1$)

Solution 2:

Here's another solution.

Observe that $$\sum \frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \sum a_i - a_{i+1} = 0, $$

Hence $\sum \frac{a_{i}^2}{a_i+a_{i+1}} = \sum \frac{a_{i+1}^2}{a_i+a_{i+1}}$, and we just need to show that

$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq 1.$$

This makes the inequality much more symmetric, and easier to manipulate. In particular,

$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq \sum \frac{1}{2} (a_i + a_{i+1}) = 1$$

Solution 3:

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

$$ \frac{a_{1}^2}{a_{1}+a_{2}} + \cdots + \frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{(a_{1} + \cdots + a_{n})^{2}}{2(a_{1}+\cdots+a_{n})} = \frac{1}{2}. $$