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.
Best Answer
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 :) .