Show: $\sum_{n\le x} \phi(n) [\frac{x}{n}] = \sum_{n \le x} \sum_{m\le \frac{x}{n}} \phi(m)$
I know the left most sum boils down to $\sum_{n\le x} n$.
If we know that $m|\frac{x}{n}$ then we know this to be true, but this does not follow directly from the sum limits, does it? All we know is that $mn \le x$, not $mn=x$.
Any suggestions/hints would be beautiful!
Take a closer look $\let\leq\leqslant$at the sum $$\sum_{n\leq x}\sum_{m\leq\frac xn}\phi(m).$$
You could ask yourself the question "How many times does $\phi(m)$ appear, for a fixed $m$?"
This is, of course, the number of $n$'s for which $m\leq\frac xn$, or equivalently $n\leq\frac xm$. By definition of the floor function there are exactly $\left\lfloor\frac xm\right\rfloor$ such $n$.
This means that this sum equals (since $m$ can be at most $x$)
$$\sum_{m\leq x}\phi(m)\cdot\left\lfloor\frac xm\right\rfloor.$$
Now we just have to replace $m$ by $n$ to obtain the desired sum.
What we essentially did is changing the order of summation:
$$\begin{align}\sum_{n\leq x}1\sum_{mn\leq x}\phi(m)&=\sum_{m\leq x}\phi(m)\sum_{mn\leq x}1\\&=\sum_{m\leq x}\phi(m)\cdot\left\lfloor\frac xm\right\rfloor.\end{align}$$
Note:
In general it is true that $$\sum_{n\leq x}\sum_{m\leq\frac xn}f(m)=\sum_{m\leq x}f(m)\cdot\left\lfloor\frac xm\right\rfloor$$ for any function $f$. We can apply exactly the same reasoning as above.