Local Mathematical Competitions

Assume by contradiction that $$\frac{x_1}{y_1^p}+\frac{x_2}{y_2^p}+\cdots +\frac{x_n}{y_n^p}<x_1+x_2+\cdots +x_n.$$ On the other hand, by multiplying the given relation by $p\in \mathbb{N}$, we have $$p(x_1+x_2+\cdots +x_n)\ge p(x_1{y}_1+x_2{y}_2+\cdots +x_n{y}_n).$$ Adding the above inequalities we get $$(p+1)\sum _{i=1}^n{x}_i>\sum _{i=1}^n{x}_i\left(p{y}_i+\frac{1}{y_i^p}\right).$$ But by the AM-GM inequality we have $$p{y}_i+\frac{1}{y_i^p}\ge p+1.$$ Finally, $(p+1){\sum }_{i=1}^n{x}_i>(p+1){\sum }_{i=1}^n{x}_i$, which is false.

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Alternative solution.

We shall use the following inequality due to Radon.

LEMMA. If $x_1,x_2,\dots ,x_n$ and $y_1,y_2,\dots ,y_n$ are positive real numbers, and $p$ is a non-negative real number, then the following inequality holds $$\frac{x_1^{p+1}}{y_1^p}+\frac{x_2^{p+1}}{y_2^p}+\cdots +\frac{x_n^{p+1}}{y_n^p}\ge \frac{(x_1+x_2+\cdots +x_n{)}^{p+1}}{(y_1+y_2+\cdots +y_n{)}^p}.$$ Proof. We can rewrite the inequality as $$\sum _{i=1}^n{y}_i{\left(\frac{x_i}{y_i}\right)}^{p+1}\ge \left(\sum _{j=1}^n{y}_j\right){\left(\frac{{\sum }_{i=1}^n{x}_i}{{\sum }_{i=1}^n{y}_i}\right)}^{p+1}$$ or equivalently $$\sum _{i=1}^n\frac{y_i}{{\sum }_{j=1}^n{y}_j}{\left(\frac{x_i}{y_i}\right)}^{p+1}\ge {\left(\frac{{\sum }_{i=1}^n{x}_i}{{\sum }_{i=1}^n{y}_i}\right)}^{p+1}.$$ Let us consider the convex function $f:[0,\infty )\to \mathbb{R}$ defined by $f(x)=x^{p+1}$. The last inequality follows from Jensen's inequality $${\lambda }_{1}f({\alpha }_1)+{\lambda }_{2}f({\alpha }_2)+\cdots +{\lambda }_{n}f({\alpha }_n)\ge f({\lambda }_1{\alpha }_1+\cdots +{\lambda }_n{\alpha }_n),$$ where ${\lambda }_i=\frac{y_i}{y_1+y_2+\cdots +y_n}>0,{\sum }_{i=1}^n{\lambda }_i=1$ and ${\alpha }_i=\frac{x_i}{y_i}$. $\square$

Now, returning to our problem, by applying the LEMMA and the hypothesis of the problem, we obtain $$\sum _{i=1}^n\frac{x_i}{y_i^p}=\sum _{i=1}^n\frac{x_i^{p+1}}{(x_i{y}_i{)}^p}\ge \frac{{\left({\sum }_{i=1}^n{x}_i\right)}^{p+1}}{{\left({\sum }_{i=1}^n{x}_i{y}_i\right)}^p}\ge \sum _{i=1}^n{x}_i.$$