Let $a,$ $b,$ $c,$ $d$ be positive real numbers.  Find the minimum value of
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right).\]
Hey, this looks like a job for the AM-HM inequality! It sure does. By AM-HM, we have:
$$\frac{a + b + c + d}{4} \geq \frac{4}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$$ It's a fancy name for a simple concept. Basically, the arithmetic mean of a set of numbers is always greater than or equal to the harmonic mean. So what you're saying is that $\frac{a+b+c+d}{4} \ge \frac{4}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$. Exactly. But we can rearrange this to get the desired result. So the minimum value of $(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$ is $4*4=16$. Right. Let's see if we can reach that minimum. Well, we can't make any of the numbers $a,b,c,d$ $0$, because they have to be positive. And that gives us $(1+1+1+1)(\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}) = (4)(4) = 16$ So the minimum value is $16$

