0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

\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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.