Prove $\sum\limits_{k=1}^n a_k\sum\limits_{k=1}^n \frac1{a_k}\le\left(n+\frac12\right)^2$ then $\max a_k \le 4\min a_k$

Question

Prove that

if \begin{align} &0<a_1,a_2,\dots,a_n \in \mathbb R, &\left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n \frac1{a_k}\right)\le\left(n+\frac12\right)^2\\ \end{align}

then $$\max_k \space a_k \le 4\times\min_k\space a_k$$

What I've tried was $$\left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n \frac1{a_k}\right)=n+\sum_{i\ne j}\frac{a_i}{a_j}=n+\sum_{i< j}\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)$$ which didn't help at all. Thanks.

Answer 1

Assume that $a_1\le\ldots\le a_n$. Suppose the contrary: $a_n>4a_1$.

Let $S$ be the sum $$S=\sum_{1\le i<j\le n}\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)$$

Let's split this sum in three "parts" to bound them below:

The first, $S_1$ is just $a_1/a_n+a_n/a_1$.

$$S_1=\frac{a_1}{a_n}+\frac{a_n}{a_1}>4+\frac14=\frac{17}4$$

$S_2$ is the sum of the fractions where occurs $a_n$ or $a_1$, but not both:

$$\begin{align} S_2&=\sum_{k=2}^{n-1}\left(\frac{a_k}{a_1}+\frac{a_k}{a_n}+\frac{a_1}{a_k}+\frac{a_n}{a_k}\right)\\ &=\left(1+\frac{a_1}{a_n}\right)\sum_{k=2}^{n-1}\left(\frac{a_k}{a_1}+\frac{a_n}{a_k}\right)\\ &\stackrel*\ge2(n-2)\left(1+\frac{a_1}{a_n}\right)\left(\frac{a_n}{a_1}\right)^{1/2}\\ &=2(n-2)\left[\left(\frac{a_n}{a_1}\right)^{1/2}+\left(\frac{a_1}{a_n}\right)^{1/2}\right]\\ &>\frac52\cdot 2(n-2)\ge 5(n-2) \end{align}$$

Last, $S_3$ is the sum of the remaining terms:

$$S_3=\sum_{2\le i<j\le n-1}\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)\ge2\binom{n-2}2=n^2-5n+6$$

So $$S>\frac{17}4+5n-10+n^2-5n+6=n^2+\frac14$$

and then $$n+S>\left(n+\frac12\right)^2$$

Remarks: Throughout the proof I have been using two facts:

• The function $f(x)=x+x^{-1}$ for $x>1$ is strictly increasing.
• The sum of $r$ positive numbers whose product is $1$ is at least $r$. (This follows easily from AM-GM inequality).

For the inequality marked with $*$ I have used AM-GM inequality, too.

Answer 2

Maybe the following reasoning can help.

Let $f(a_1,a_2,...a_n)=\sum\limits_{k=1}^n a_k\sum\limits_{k=1}^n \frac1{a_k}$, $\max_{i}a_i=M$, $\min_{i}a_i=m$ and $M=xm$.

We need to prove that $x\leq4$.

Indeed, let $x>4$.

Easy to see that $\frac{\partial^2f}{\partial a_i^2}>0$ for all $i$, which says that $\max\limits_{m\leq a_i\leq M}f=\max\limits_{a_i\in\{m,M\}}f$.

Let $m$ gotten $k$ times.

Hence, $(km+(n-k)M)\left(\frac{k}{m}+\frac{n-k}{M}\right)\leq\left(n+\frac{1}{2}\right)^2$ or $k^2+(n-k)^2+k(n-k)\left(x+\frac{1}{x}\right)\leq\left(n+\frac{1}{2}\right)^2$ and since $x+\frac{1}{x}>\frac{17}{4}$, we obtain $$9k(n-k)<4n+1,$$ which is strange.

Answer 3

Another way: When you see "reversals" of CS inequality, often Pólya-Szegö’s inequality is useful.

With $\min a_k = m, \max a_k = Cm$, using Pólya-Szegö, we have $$\sum a_k \sum \frac1{a_k} \geqslant \frac{n^2}4\left(C +\frac{1}C \right)^2$$

$$\therefore \frac{n}2\left(C + \frac1C\right) \leqslant n+\frac12 \implies C + \frac1C \leqslant 2 + \frac1n < 3 \implies C < \frac12(1+\sqrt5) < 4$$ We can strictly bound this as for $n=1$ trivially $C=1$. In fact the above establishes $C < 1.62$.

Answer 4

Hint: \begin{align*} \min_j a_j &\leq a_k \quad \text{ for every } 1 \leq k \leq n\\ \sum_{k = 1}^n \min_j a_j &\leq \sum_{k = 1}^n a_k\\ n \, \min_j a_j &\leq \sum_{k = 1}^n a_k \end{align*} Similarly $$ n \frac{1}{\max\limits_j a_j} \leq \sum_{k = 1}^n \frac{1}{a_k} $$ Hence $$ n^2 \frac{\min\limits_j a_j}{\max\limits_i a_i} \leq \left( \sum_{k = 1}^n a_k \right)\left( \sum_{k = 1}^n \frac{1}{a_k} \right) \leq \left( n + \frac{1}{2} \right)^2 \leq \ldots $$ This should lead to something. Also, obviously $\frac{\max\limits_i a_i}{\min\limits_j a_j} \geq 1$. (I like having different indeces $i, j$ but these are of course dummy variables)