Arithmetic Mean/Geometric Mean Inequality We have to use the arithmetic mean/geometric mean inequality.
Let $a$ and $b$ be fixed positive numbers. Find the value of $x>0$ that minimizes the given expression and determine the minimum value of the expression,
$ax+\frac{b}{x}$.
I don’t know how to get started by transforming the expression to be able to use the mean inequality. We were told not to solve using calculus.
 A: The $AM-GM$ inequality tells us that for non-negative $a$ and $b$ we have
$$\frac{a+b}{2}\geq \sqrt{ab}$$
Since $x>0$ then $ax>0$ and by $AM-GM$
$$\frac{ax+\frac{b}{x}}{2}\geq\sqrt{ax\frac{b}{x}}$$
Or $$ax+\frac{b}{x}\geq2\sqrt{ab}$$
with equality when $ax+\frac{b}{x}=2\sqrt{ab}$. Can you end it now?
A: Ares has provided an excellent algebraic solution that certainly answers your question. Based on how this site is supposed to work, that answer is better than mine. But I would like to also provide an answer for you that does not jump straight into an algebraic solution; but takes a moment to practice some important mathematical skills for problem solving which come in handy when we are stumped! It is a useful skill in math to learn how to explore a problem if we don't know where to begin on a solution.
What will the function $f(x) = ax + b/x$ look like? Are we convinced it has a minimum?

*

*When $x$ is small $ax$ is going to be small and not contribute much to the function and $b/x$ will be large and dominate the function. For small $x$ we can expect the function to approximately look like $b/x$, which for any positive value of $b$ will look approximately like the function $1/x$ which is a decreasing function.


*As $x$ grows, $b/x$ will start to get small and contribute less to the function and $ax$ will grow and begin to dominate the function. As $x$ gets very large the function will get closer and closer to the function $ax$, which for all positive values of $a$ is an increasing linear function.


*From our work above we expect that for all positive values $a,b$ our function will switch from decreasing to increasing which guarantees a minimum.
The function out to look something like this:

If you really want to play around with graphing this function for intuition, you could make a tool like this on desmos which allows you to toggle the values of $a,b$ (though with more practice I don't think this is necessary)

Where might the minimum occur?
We know the minimum should be wherever the function "switches" from $b/x$ to $ax$, but this will depend on the chosen values of $a,b$.

*

*If $a$ is large, the function will switch to $ax$ sooner because $ax$ will grow faster. This makes the minimizer smaller.


*If $b$ is large, the function will switch to $ax$ longer because it will take a larger $x$ to make $b$ small. This will make the minimizer bigger.
Moving to the algebra
At this point I am convinced of the existence of the minimum, and I am convinced it will be some function of the numbers $a,b$. I feel more grounded in the problem. I know I need to use the AM-GM inequality which in its simple case states that for any nonnegative $a,b \in \mathbb{R}$:
$$\frac{a+b}{2} \geq \sqrt{ab}.$$
Since $x > 0, $a > 0$, $b > 0$ then $ax$ and $b/x$ are both positive real numbers. At this point, the only logical thing to try is top plug the numbers into the inequality to yield
$$\frac{ax + \frac{b}{x}}{2} \geq \sqrt{ax\frac{b}{x}}, \hspace{2mm} \text{ for all} x \in \mathbb{R}$$,
The part about "for all $x \in \mathbb{R}$" is very important!
At this point you can pick up where Ares solution starts. Good luck!
