<s>[INST] $E(X)E(1/X) \leq (a + b)^2 / 4ab$ I've worked on the following problem and have a solution (included below), but I would like to know if there are any other solutions to this problem, especially more elegant solutions that apply well known inequalities.
QUESTION: Suppose we have a random variable s.t. $P(a<X<b) =1$  where 0 < a < X < b , a and b both positive constants.
Show that $E(X)E(\frac{1}{X}) \le \frac{(a+b)^2}{4ab}$
Hint:  find constant c and d s.t. $\frac{1}{x} \le cx+d$ when $a<x<b$, and argue that then we shall have $E(\frac{1}{X}) \le cE(X)+d$
MY SOLUTION: For a line $cx+d$ that cuts through $\frac{1}{X}$ at the points x=a and x = b, it's easy to show that $ c = - \frac{1}{ab} $  and $d = \frac{a+b}{ab} $, 
So, $ \frac{1}{X} \le - \frac{1}{ab} X + \frac{a+b}{ab} $, and therefore:
$$ E(\frac{1}{X}) \le - \frac{1}{ab} E(X) + \frac{a+b}{ab} $$  
$$ abE(\frac{1}{X}) + E(X) \le (a+b) $$
Now, because both sides of the inequality are positive, it follows that:
$$ [abE(\frac{1}{X}) + E(X)]^2 \le (a+b)^2 $$
$$ (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2  \le (a+b)^2 $$ 
Then, for the LHS, we can see that 
$2abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + E(X)^2$ 
because  
$0 \le (ab)^2E(\frac{1}{X})^2 - 2abE(\frac{1}{X})*E(X) + E(X)^2 = [abE(\frac{1}{X}) - E(X)]^2 $
SO,
$$ 4abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2  \le (a+b)^2 $$
and therefore:
$$ E(\frac{1}{X})E(X) \le \frac{(a+b)^2}{4ab} $$  Q.E.D.
Thanks for any additional solutions you might be able to provide.  Cheers!
 [/INST] I know it's stated in the problem, but I figured I'd put it in the answer bank:
For some line $cx+d$ that cuts through $\frac{1}{X}$ at the points x=a and x = b, it's easy to show that $ c = - \frac{1}{ab} $  and $d = \frac{a+b}{ab} $, 
So, $ \frac{1}{X} \le - \frac{1}{ab} X + \frac{a+b}{ab} $, and therefore:
$$ E(\frac{1}{X}) \le - \frac{1}{ab} E(X) + \frac{a+b}{ab} $$  
$$ abE(\frac{1}{X}) + E(X) \le (a+b) $$
Now, because both sides of the inequality are positive, it follows that:
$$ [abE(\frac{1}{X}) + E(X)]^2 \le (a+b)^2 $$
$$ (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2  \le (a+b)^2 $$ 
Then, for the LHS, we can see that 
$2abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + E(X)^2$ 
because  
$0 \le (ab)^2E(\frac{1}{X})^2 - 2abE(\frac{1}{X})*E(X) + E(X)^2 = [abE(\frac{1}{X}) - E(X)]^2 $
SO,
$$ 4abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2  \le (a+b)^2 $$
and therefore:
$$ E(\frac{1}{X})E(X) \le \frac{(a+b)^2}{4ab} $$
</s>