Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$

Question

For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that $$ (a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2. $$ Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.

Can someone help me [rove this? It's USAMO 2009 .I thought of using cauchy.

Answer 1

W.L.O.G. Assume $a_{1}$ is the minimum and $a_{2}$ is the maximumm if we have $x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n} \in \mathbb R $ then by cauchy-Schwars we have $(x_{1}y_{1}+...+x_{n}y_{n})^2\leq(x_{1}^2+...+x_{n}^2)(y_{1}^2+...+y_{n}^2)$, we need an equation relating $a_{1}$ to $a_{2}$, if we put $\sqrt a_{2}, \sqrt a_{1}, \sqrt a_{3},....,\sqrt a_{n}$ in place of $x_{1},....x_{n}$ and $\frac{1}{\sqrt a_{1}},....,\frac{1}{\sqrt a_{n}}$ in place of $y_{1},...,y_{n}$ this gives $(\sqrt \frac{a_{2}}{a_{1}}+\sqrt \frac{a_{1}}{a_{2}}+n-2)^2 \leq (a_{2}+a_{1}+...+a_{n})(\frac{1}{a_{1}}+\frac{1}{a_{2}}+....+\frac{1}{a_{n}})$ combining this with the above inequality it will be much easier to proceed :) .

Answer 2

Update:

Just realized we can prove directly without taking derivative.

WLOG assume $a_1 \ge a_2 \ge \cdots \ge a_n$, and $a_1 > 4a_n$ then

$$\begin{split} & & \left(\sum a_i \right) \left( \sum \frac{1}{a_i} \right) \\ & = & n + \left(\frac{a_1}{a_n}+\frac{a_n}{a_1}\right) + \sum_{1 < i<j<n} \left( \frac{a_i}{a_j} + \frac{a_j}{a_i}\right) + \sum_{i=2}^{n-1} \left( \frac{a_i}{a_1} + \frac{a_n}{a_i} \right) + \sum_{i=2}^{n-1} \left( \frac{a_i}{a_n} + \frac{a_1}{a_i} \right)\\ & > & n+\left(4+\frac 14\right) + 2 \binom{n-2}{2} + (n-2)\left(2\sqrt{\frac{a_n}{a_1}}+ 2\sqrt{\frac{a_1}{a_n}}\right)\\ & > & n + 4 + \frac 14 + (n-2)(n-3) + 2(n-2) \left( 2 + \frac 12 \right)\\ & = & n + 4 + \frac 14 + (n-2)(n+2) = n^2 + n + \frac 14 \left(n+\frac 12\right)^2 \\ \end{split}$$

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Original proof:

We prove by induction that if $\max(a_i) > 4 \min (a_i)$ then $\sum a_i \sum \frac{1}{a_i} > \left(n+\frac 12\right)^2$

When $n=2$ it's true.

Suppose it's true for $n$. WLOG we assume $a_1 \ge a_2 \ge \cdots \ge a_n$ and $a_{n+1}$ is between $a_n$ and $a_1$. Denote

$$f(x)=\frac{a_1}{x} + \frac{x}{a_1} + \frac{a_n}{x}+\frac{x}{a_n}$$

$$f^{(2)}(x) > 0,f'(x)=\frac{1}{a_1} + \frac{1}{a_n} - \frac{a_1+a_n}{x^2}$$

It's easy to show that $f(x)$ obtains its minimum at $x=\sqrt{a_1 a_n}$, and $$f(\sqrt{a_1a_n})=2 \left(\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}} \right) > 2(2+\frac 12)=5$$

Define $S_n = \sum_{i=1}^n a_n, T_n = \sum_{i=1}^n 1/a_n$. Then $$ \sum_{i=1}^{n+1} a_i \sum_{i=1}^{n+1} \frac{1}{a_i} = (S_n+a_{n+1})\left(T_n+\frac{1}{a_{n+1}}\right) \\ = S_n T_n + 1 + \sum_{i=2}^{n-1} \left(\frac{a_{n+1}}{a_i} + \frac{a_i}{a_{n+1}}\right)+f(a_{n+1})\\ > \left(n+\frac 12\right)^2 + 1 + 2(n-2)+5=\left(n+1+\frac 12\right)^2.\blacksquare $$