$E(X)E(1/X) \leq (a + b)^2 / 4ab$ 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!
 A: 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:
$$ E(\frac{1}{X})E(X) \le \frac{(a+b)^2}{4ab} $$
A: Once we observe that both sides of the inequality are scale invariant, it follows immediately by combining two simple, well-known inequalities, of which the most notable is that correlation coefficients cannot be less than $\newcommand{\e}{\mathbb{E}}-1$.

The Cauchy-Schwarz Inequality guarantees that the correlation coefficient $\rho(X,Y)$ of any two random variables $X$ and $Y$ lies between $-1$ and $1$.  Using the definition of correlation and focusing on the lower bound of $-1$ allows us to express this inequality in the form
$$-\text{sd}(X)\text{sd}(Y) \le \rho(X,Y)\text{sd}(X)\text{sd}(Y) = \text{Cov}(X,Y) = \e[XY] - \e[X]\e[Y].$$
If we let $Y=1/X$, the product $\e[X]\e[1/X]$ is recognizable right there at the very end (which of course is what inspired this approach).
Using the simplification $\e[XY]=\e[X/X]=\e[1]=1$, isolate the last term algebraically to obtain
$$\e[X]\e[1/X] \le 1 + \text{sd}(X)\text{sd}(1/X).$$
When a variable's values are confined to an interval $[a,b]$, its variance is limited by the value $(b-a)^2/4$.  This is proven in several elegant, elementary, and informative ways at Variance of a bounded random variable; it comes down to the fact that variances cannot be negative. Consequently
$$\text{sd}(X)\text{sd}(1/X) \le (b-a)^2/4.$$
We may freely rescale $X$ because the left side of this inequality does not thereby change.  Select a scale in which $ab=1$ (which is possible because both $a$ and $b$ are positive).  This allows us to rewrite the preceding in a form where the right hand side is obviously scale invariant, too:
$$\text{sd}(X)\text{sd}(1/X) \le \frac{(b-a)^2}{4} = \frac{1}{ab}\frac{(b-a)^2}{4}=\frac{(b-a)^2}{4ab}.$$
These two inequalities finish the job:
$$\e[X]\e[1/X] \le 1 + \text{sd}(X)\text{sd}(1/X) \le 1 + \frac{(b-a)^2}{4ab} = \frac{(a+b)^2}{4ab}.$$
