Proof of minimum value of x+y when xy = a Suppose that a is a positive integer and that x and y are positive real numbers such that x*y = a. Then what is the minimum value of (x+y)?
Intuitively, it seems obvious that the minimum positive value of x + y would exist when both x and y are equal to the positive square root of a. But if this is true how would you prove it? And, if not how can the minimum value of x + y be found?
 A: Hint: We have
$$(x+y)^2=4xy+(x-y)^2=4a+(x-y)^2.$$
The right side is smallest when $x-y=0$.
A: You are looking for the minimum value of $x+y$ when $xy=a$. From the last, you can extract $y=\frac{a}{x}$ and then $$f(x)=x+\frac{a}{x}$$ is the function you want to minimize.By derivation, we have $$f'(x)=1-\frac{a}{x^2}$$  $$f''(x)=\frac{2a}{x^3}$$ The first derivative cancels for $x=\sqrt a$ and, for this value, the second derivative is positive, then $x=y=\sqrt a$ corresponds to the minimum value.
A: Here's one more solution:
Let y = a/x. One can write the sum as a function of x: $f(x) = x + \frac{n}{x}$. We are looking for the minimum value of this function on the positive reals. Since $f(x) > 0$ for each x > 0, let $\alpha$ be the minimum value, we can rewrite the equation as $$x + \frac{n}{x} - \alpha = 0$$. Multiplying by x: $$x^2 - \alpha x + n = 0$$. This is a quadratic equation for x, and since we are looking for the mininum value of alpha, it touches the x axis only once, which means that the discriminant is zero. $$\alpha^2 - 4n = 0$$ and the solution follows.
