Minimum value of the inequality

Question

Give $a_{1}\geq a_{2}\geq ...\geq a_{n}> 0$ and a positive integer m . Find the minimum value of the following the inequality:

$\left ( a_{1}+a_{2}+...+a_{n} \right )\left ( \frac{1}{a_{1}^{m}}+ \frac{1}{a_{2}^{m}}+...+\frac{1}{a_{n}^{m}}\right )$

Please help me to solve the above problem. I try but not success. Thank you very much.

Answer 1

Let us define $$ F_m(a_1,\ldots,a_n)=\left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^na_k^{-m}\right). $$

• If $m>1$ we have $$\lim_{x\to\infty}F_m(x,\ldots,x)=\lim_{x\to\infty}(n^2x^{1-m})=0$$ So, $\inf F_m=0$, in this case. (Note that the minimum in not attained here).
• If $m=1$, then by th Cauchy schwarz inequality we have $$n^2=\left(\sum_{k=1}^na_k\cdot \frac{1}{a_k}\right)^2\leq \left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^na_k^{-1}\right)=F_1(a_1,\ldots,a_n) $$ with equality if $a_1=a_2=\ldots=a_n$. So, $\min F_1=n^2~$ in this case.