Proving $\forall a\;\exists b\;\forall c,d$ such that $|c-d|≤b ∨|c^2-d^2|>a$ (with all values positive reals)

Question

I'm interested in proving the following statement:

$$\forall a \in \mathbb{R}^+,\exists b\in \mathbb{R}^+,\forall c,d\in\mathbb{R}^+,|c-d|≤b ∨|c^2-d^2|>a$$

I haven't made much ground as all I can think of doing is factoring out the second absolute value:

$$|c^2-d^2|=|c-d||c+d|$$

Then maybe do something with cases, $|c-d|≤b$ and $|c-d|>b$ where the second case results in:

$$|c-d||c+d|>(b)|c+d|$$

Is this the right approach? Otherwise, how should I go about doing it?

Answer 1

It's seems to be most intuitive to use the equivalence $p\lor q \equiv (\neg p \to q)$ to rewrite the goal into $$(\forall a \in \mathbb{R}^+)(\exists b\in \mathbb{R}^+)(\forall c,d\in\mathbb{R}^+)\bigl (|c-d|>b \to |c^2-d^2|> a\bigr)$$

So for each $a$ you're asked to find a $b$ such that increasing any positive $x$ by $b$ will guarantee that $x^2$ increases by at least $a$.

Looking at a graph for $x^2$ it seems plausible that $b=a$ will work, unless the initial $x$ is so small that $\frac{d}{dx}x^2<1$. But we force the increase in $x$ to get us out of that area simply by adding a sufficiently large constant to $b$. In other words, $b=\frac12+a$ ought to work.

And now you can begin looking for an algebraic argument that this choice of $b$ actually does work.

Answer 2

Let consider two cases

• $c=d\implies |c-d|=0\le1=b$
• $c\neq d\implies |c-d|=k\le k=b$