# A calculus of the absurd

##### 4.12.4 Some interesting inequality proofs
• Example 4.12.2 Let $$a, b, c$$ be positive reals such that

$$a^2 - b^2 > c^2 - a^2.$$

Prove that

$$a - b > c - a$$

We want to prove the statement

\begin{align} a^2 - b^2 > c^2 - a^2 \implies a - b > c - a. \end{align}

We can prove the equivalent statement,

$$a - b \leqq c - a \implies a^2 - b^2 \leqq c^2 - a^2.$$

Which in turn is the same as

$$2a \leqq b + c \implies 2a^2 \leqq c^2 + b^2.$$

Let us fix some $$a, b, c \in \mathbb {R}$$ such that $$a, b, c \geqq 0$$. Then let us assume further that $$2a \leqq b + c$$ in which case

\begin{align} 2a^2 &\leqq 2\left (\frac {b+c}{2}\right )^2 \\ &\leqq \frac {1}{2}c^2 + cb + \frac {b^2}{2} \end{align}

This is pretty close to what we want - in fact if we could prove that $$cb \leqq \frac {1}{2}c^2 + \frac {1}{2}b^2$$ we would be done.

\begin{align} & cb \leqq \frac {1}{2}c^2 + \frac {1}{2}b^2 \\ &\iff 0 \leqq \frac {1}{2}c^2 - cb + \frac {1}{2} b^2 \\ &\iff 0 \leqq c^2 - 2cb + b^2 \\ &\iff 0 \leqq (c - b)^2 \end{align}

and the last statement is always true.

Therefore,

\begin{align} 2a^2 &\leqq \frac {1}{2}c^2 + cb + \frac {b^2}{2} &\leqq \frac {1}{2}c^2 + \frac {1}{2}c^2 + \frac {1}{2}b^2 + \frac {b^2}{2} \\ &\leqq c^2 + b^2 \end{align}

from which it follows that

\begin{align} a^2 - b^2 \leqq c^2 - a^2 \end{align}

and then we can forget that we fixed $$a, b, c$$ and we have a proof for any $$a, b, c$$, so we are done.

$$\Box$$