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.