Proof with inequalities and a function I need help approaching a proof which deals with inequalities:

If p and r are the precision and recall of a test, then the F1 measure of the test is
defined to be
$$F(p, r) = \frac{2pr}{p+r}$$

Prove that, for all positive reals p, r, and t, if t ≥ r then F(p, t) ≥ F(p, r)

What's the first step to approaching this problem? Do I need to look at this with different cases?
A standard way is to look at $F(p,t)-F(p,r)$, which is
$$\frac{2pt}{p+t}-\frac{2pr}{p+r}.$$
Bring to a common denominator and simplify. We get
$$\frac{2pt(p+r)-2pr(p+t)}{(p+t)(p+r)},$$
which simplifies to
$$\frac{2p^2(t-r)}{(p+t)(p+r)}.$$
This is clearly $\ge 0$.
$$\frac{1}{p}+\frac{1}{r}=\frac{p+r}{pr}\\so\\ f(p,r)=2 \frac{1}{\frac{1}{p}+\frac{1}{r}}\\\frac{1}{\frac{1}{p}+\frac{1}{r}}\\\\is -harmonic-mean$$