# Proofs: Problem 4

### Model Solution

Here is one fairly standard way to do this proof. Notice that much of the second paragraph will come as a bit of a surprise to the reader. The scratch work was done by working backwards from the conclusion, then we reversed it to create the formal proof.

Assume that p,t, and r are positive reals, and that $$t \geq r$$.

Multiplying by $$2p^2$$ (which is positive, so the inequality doesn't change), we get $$2p^2t \geq 2p^2r$$. This implies that $$2p^2t + 2ptr \geq 2p^2r + 2ptr$$ or equivalently that $$(2pt)(p+r) \geq (2pr)(p+t)$$.

Now dividing each side by the positive numbers (p+r) and (p+t), we get $$\frac{2pt}{p+t} \ge \frac{2pr}{p+r}$$. By the definition of F, this means that $$F(p,t) \geq F(p,r)$$, which was what we needed to prove.

This style of proof is straightforward to produce, once you've done the backwards scratch work.

### Alternate Proof

The proof above seems unnatural, because the reader won't immediately see why we should be multiplying by $$2p^2$$. Here is an alternative proof which takes more effort from the writer but is easier for the reader.

Let p, r, and t be positive real numbers. Suppose that $$t \ge r$$. To prove $$\frac{2pt}{p+t} \ge \frac{2pr}{p+r}$$, it is sufficient to show that $$\frac{2pt}{p+t} - \frac{2pr}{p+r} \ge 0$$.

Now, we can compute $$\begin{eqnarray*} \frac{2pt}{p+t} - \frac{2pr}{p+r} & = &\frac{(2pt)(p+r) - (2pr)(p+t)}{(p+t)(p+r)} \\ & = & \frac{2p^2t+2ptr - 2p^2r-2ptr}{(p+t)(p+r)} \\ & = & \frac{2p^2t-2p^2r}{(p+t)(p+r)} \\ & = & \frac{2p^2(t-r)}{(p+t)(p+r)} \\ & \geq & 0\\ \end{eqnarray*}$$

The last step holds because $$(t-r)$$, $$p^2$$, $$p+t$$ and $$p+r$$ are all positive real numbers.

Since $$\frac{2pt}{p+t} - \frac{2pr}{p+r} \ge 0$$, we know that $$\frac{2pt}{p+t} \ge \frac{2pr}{p+r}$$, which is what we needed to show.