# 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?

Hint: We have $$(x+y)^2=4xy+(x-y)^2=4a+(x-y)^2.$$ The right side is smallest when $x-y=0$.
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.