Sum greater than 1; minimization for non-strict inequalities

I want to show that if $x_k>0$ for $k=1,2,...,n$ and $\sum_{k=1}^n x_k=1$, then $\sum \frac{x_k^2}{y_k}\ge 1$ for any $y_1, y_2,...,y_n>0$ so that $\sum_{k=1}^n y_k=1$.

I tried solving the minimization problem $\min \sum_{k=1}^n \frac{x_k^2}{y_k}$ $s.t. p_k>0, \sum_{k=1}^n p_k=1$. But I'm not sure how to solve such a problem since there are non-strict inequalities in the constraints.

So I am hoping for:

(1) Any pointers as to proving this inequality $\sum \frac{x_k^2}{y_k}\ge 1$ , and

(2) References/tips for a general method to solve such non-strict inequality minimization problems.

Thanks!

1 Answer

By the Cauchy-Schwarz inequality,

\begin{align*} 1 &= \left(\sum_{k = 1}^n x_k\right)^2 \\ &= \left(\sum_{k = 1}^n \frac {x_k} {\sqrt{y_k}} \sqrt{y_k}\right)^2\\ &\le \left(\sum_{k = 1}^n \frac{x_k^2}{y_k}\right) \cdot \left(\sum_{k = 1}^n y_k\right) \\ \end{align*}

from which the result follows.

The fact that a) the terms can naturally be recognized as squares and b) we have specific information about the sums of non-squared terms together suggested this route. Had the powers not been $2$, then we could perhaps have used something like Holder's inequality.

• Excellent, thanks Mr. Bongers! – manofbear Apr 24 '16 at 16:37