# Can I use induction?

1. May 19, 2014

### knowLittle

1. The problem statement, all variables and given/known data
Prove for any $a,b \in \mathbb{R^+} : |a-b| \leq \sqrt{a^2 +b^2}$

3. The attempt at a solution

How should I start? Can I use induction?
Should I use contrapositive?

2. May 19, 2014

### Curious3141

You're overthinking it. Consider $(|a-b|)^2$ first.

3. May 19, 2014

### knowLittle

uh, are you serious?

$(|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2$

Then the LHS will always be lesser for any $\mathbb{R^+}$

4. May 19, 2014

### Curious3141

Pretty much, except that going from this step to this: $(a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2$ is unnecessary. You can just go from this to this: $(a^2 - 2ab + b^2 ) \leq a^2 +b^2$ by observing that $2ab > 0$.

As a final step, you still have to take the square root of both sides. You have to justify that the inequality doesn't change direction by observing the monotonic increasing nature of the square root function (just a simple sketch would do).

BTW, if a and b are both positive reals, the strict inequality sign can be used (a stronger statement), i.e. $|a-b| < \sqrt{a^2 + b^2}.$

Last edited: May 19, 2014
5. May 19, 2014

### knowLittle

Wouldn't I need this first:

$-\sqrt{ a^2 + b^2} < (a-b) < \sqrt{ a^2 +b^2 }\\$

Then,
$a^2 +b^2 < ( a-b )^2 = (a^2 - 2ab + b^2 ) < (\sqrt{a^2 +b^2})^2 = a^2 +b^2$
Now, I don't understand

6. May 19, 2014

### Curious3141

I'm sorry, I do not understand what you're doing here.

The first step is to consider the square of the expression $|a-b|$.

It should be trivial to recognise that $(|a-b|)^2 = |(a-b)^2| = (a-b)^2 = a^2 + b^2 - 2ab$

The next step is to note that $a^2 + b^2 - 2ab < a^2 + b^2$ for positive reals a and b.

That gives you $(|a-b|)^2 < a^2 + b^2$. The final step is to take the square root of both sides.

Can you finish it off from here?

7. May 19, 2014

### knowLittle

I see what you are doing.

About what I wrote prior to this...
$|a-b| < c$, then
$-c< a-b <c$

8. May 19, 2014

### Curious3141

That is true, but not relevant here.