Inequality

For all $x_1, x_2, \dots, x_n \geq 0$ , let $x_{n+1}=x_1$. Prove that:

$$\sum_{k=1}^{n}\sqrt{\frac{1}{\left ( x_k+1 \right )^2} + \frac{x_{k+1}^2}{\left ( x_{k+1}+1 \right )^2}} \geq \frac{n}{\sqrt{2}}$$

Solution

We begin by the basic inequality;

$$\sqrt{a^2+b^2} \geq \frac{a+b}{\sqrt{2}}$$

Applying this identity we have that:

\begin{align*}
\sum_{k=1}^{n}\sqrt{\frac{1}{\left ( x_k+1 \right )^2} + \frac{x_{k+1}^2}{\left ( x_{k+1}+1 \right )^2}} &\geq \frac{1}{\sqrt{2}}\sum_{k=1}^{n}\left [ \frac{1}{x_k+1} + \left ( 1- \frac{1}{x_{k+1}+1} \right ) \right ] \\
 &=\frac{n}{\sqrt{2}}
\end{align*}