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Prove that for all real numbers $x_1,...,x_n$ the following inequality holds: $$\sqrt{x_1^2+(1-x_2)^2}+...+\sqrt{x_{n-1}^2+(1-x_n)^2}+\sqrt{x_n^2+(1-x_1)^2} \ge \frac{n\sqrt{2}}{2}$$

I tried Induction on n, the base case worked, but for the induction step it ended in finding a minimum with positive Hesse-matrix(the problem was, that the Hesse-matrix was not positive. So I think, Induction is not working. The problem is how to manipulate the inequality directly, since it is difficult with the sums in the roots.

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2 Answers 2

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Hint: Let $s=\sum x_i$. By Minkowsky’s inequality, LHS $\geqslant \sqrt{s^2+(n-s)^2}$. Now use CS inequality $(n^2+(n-s)^2)(1+1)\geqslant n^2$ to finish.

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\begin{eqnarray*} \sqrt{\frac{\alpha^2+\beta^2}{2}} \geq \frac{\alpha+\beta}{2}. \end{eqnarray*} \begin{eqnarray*} \sqrt{x_1^2+(1-x_2)^2}+...+\sqrt{x_{n-1}^2+(1-x_n)^2}+\sqrt{x_n^2+(1-x_1)^2} \geq \frac{1}{\sqrt{2}}( x_1+1-x_2+x_2+1-x_3+\cdots+x_n+1-x_1)= \frac{n}{\sqrt{2}}. \end{eqnarray*}

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