Inequalities 4

Relevant wiki : Titu's Lemma

Solution:

By Titu's lemma,

$\begin{aligned}\large\left(\dfrac{1^2}{a_1}+\dfrac{1^2}{a_2}+\cdots\dfrac{1^2}{a_{37}}\right)&≥\large\left(\dfrac{(37)^2}{a_1+a_2+\cdots+a_{37}}\right)\\ \large (a_1+a_2+\cdots+a_{37})\left(\dfrac{1^2}{a_1}+\dfrac{1^2}{a_2}+\cdots\dfrac{1^2}{a_{37}}\right)&≥\large\left(\dfrac{(37)^2}{\cancel{a_1+a_2+\cdots+a_{37}}}\right)\cancel{(a_1+a_2+\cdots+a_{37})}\\ \\ K&=(37)^2\\&=\boxed{1369}\end{aligned}$

General result:

As $\{a_i\}$ is a positive real number, therefore, we can apply AM-HM inequality as follows

$\dfrac{1}{n} \cdot \sum_{i=1}^n a_i \ge \dfrac{n}{\displaystyle \sum_{i=1}^n \frac{1}{a_i}}$

This gives

$\left( \sum_{i=1}^n a_i \right) \left( \sum_{i=1}^n \dfrac{1}{a_i} \right) \ge n^2$

Thus $X = n^2$ and putting $n=37$ gives $X = \boxed{1369}$ .

By Cauchy-Schwarz on $\big\{ \sqrt{a_1} , \sqrt{a_2} , \sqrt{a_3} , \cdots , \sqrt{ a_{37}} \big \}$ and $\big\{ \dfrac{1}{\sqrt{a_1}}, \dfrac{1}{\sqrt{a_2}}, \dfrac{1}{\sqrt{a_3}}, \cdots, \dfrac{1}{\sqrt{a_{37}}} \big\}$ , we have

$\begin{aligned} \left( a_1 + a_2 + a_3 + \cdots + a_{37} \right) \left( \dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} + \cdots + \dfrac{1}{a_{37}} \right) & \geq \left( \underbrace{ 1+1+1+\cdots+1+1 }_{37 \text{times}} \right)^2 \\ \left( \large\displaystyle \sum_{i= 1}^{37} a_i \right) \left( \large\displaystyle \sum_{i=1}^{37} \dfrac{1}{a_i} \right) & \geq 37^2 = 1369 \end{aligned}$