USAMO, 2009

Solution 1. The Cauchy-Schwarz Inequality gives
$$\begin{aligned} \left(n + \frac{1}{2}\right)^2 &\ge (a_1 + a_2 + \dots + a_n) \left(\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}\right) \\ &= (m + a_2 + \dots + a_{n-1} + M) \left(\frac{1}{M} + \frac{1}{a_2} + \dots + \frac{1}{a_{n-1}} + \frac{1}{m}\right) \\ &\ge \left(\sqrt{\frac{m}{M}} + \underbrace{1 + \dots + 1}_{n-2} + \sqrt{\frac{M}{m}}\right)^2. \end{aligned}$$
Hence
$$n + \frac{1}{2} \ge \sqrt{\frac{m}{M}} + n - 2 + \sqrt{\frac{M}{m}} \quad \text{or} \quad \sqrt{\frac{m}{M}} + \sqrt{\frac{M}{m}} \le \frac{5}{2} \qquad (2)$$
It follows that
$$2(m + M) \le 5\sqrt{Mm},$$
which is (1), completing our proof.

Solution 2 (By Adam Hesterberg). Consider the quadratic polynomial (in $x$)
$$\begin{aligned} p(x) &= \frac{1}{2} \left[ \left( \sqrt{a_1}x + \frac{1}{\sqrt{a_n}} \right)^2 + \left( \sqrt{a_n}x + \frac{1}{\sqrt{a_1}} \right)^2 + \sum_{i=2}^{n-1} \left( \sqrt{a_i}x + \frac{1}{\sqrt{a_i}} \right)^2 + \left( 5 - 2\sqrt{\frac{m}{M}} - 2\sqrt{\frac{M}{m}} \right) x \right] \\ &= \left( \frac{1}{2} \sum_{i=1}^{n} a_i \right) x^2 + \frac{2n+1}{2} \cdot x + \left( \frac{1}{2} \sum_{i=1}^{n} \frac{1}{a_i} \right). \end{aligned}$$
Its discriminant is equal to
$$\Delta = \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{i=1}^{n} \frac{1}{a_i} \right) - \left( n + \frac{1}{2} \right)^2,$$
which is nonnegative, by the given condition. Thus $p(x)$ has a root $r$, implying that
$$0 = p(r) \geq \left(5 - 2\sqrt{\frac{m}{M}} - 2\sqrt{\frac{M}{m}}\right) r.$$
Because all of $p(x)$'s coefficients are positive, we must have $r < 0$, from which it follows (2).

Solution 3 (By Zuming Feng). We set $a = \frac{a_2+\cdots+a_{n-1}}{n-2}$. Then $m \le a_2 \le a \le a_{n-1} \le M$ and $a_2 + \cdots + a_{n-1} = (n-2)a$. By the AM-HM Inequality, we have
$$\frac{1}{a_2} + \cdots + \frac{1}{a_{n-1}} \ge \frac{(n-2)^2}{a_2 + \cdots + a_{n-1}} = \frac{n-2}{a}.$$
It follows that
$$\begin{aligned} \left(n + \frac{1}{2}\right)^2 &\ge (a_1 + a_2 + \cdots + a_n) \left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}\right) \\ &\ge (m + (n-2)a + M) \left(\frac{1}{m} + \frac{n-2}{a} + \frac{1}{M}\right) \\ &= (m + M) \left(\frac{1}{m} + \frac{1}{M}\right) + (n-2)^2 + \frac{(n-2)(m+M)}{a} + (n-2)a \left(\frac{1}{m} + \frac{1}{M}\right) \\ &= \frac{(m+M)^2}{mM} + (n-2)^2 + \frac{(n-2)(m+M)}{mM} \cdot \left(\frac{mM}{a} + a\right). \end{aligned}$$
By the AM-GM Inequality, we have $\frac{mM}{a} + a \ge 2\sqrt{mM}$ with equality at $m \le a = \sqrt{mM} \le M$. We deduce that
$$\left(n + \frac{1}{2}\right)^2 \ge \frac{(m + M)^2}{mM} + (n - 2)^2 + \frac{2(n - 2)(m + M)}{\sqrt{mM}}.$$
Setting $t = \frac{m+M}{\sqrt{mM}}$ in the last inequality yields
$$\left(n + \frac{1}{2}\right)^2 \ge t^2 + (n - 2)^2 + 2(n - 2)t,$$
from which it follows that
$$t^2 + 2(n-2)t - 5n + \frac{15}{4} \le 0 \quad \text{or} \quad 4t^2 + 8(n-2)t - 5(4n-3) \le 0.$$
Factoring the left-hand side of the last inequality gives $(2t-5)(2t+(4n-3)) \le 0$, implying $2t-5 \le 0$, which is (1).