Proving |ab| ≤ (a² + b²)/2 for all real numbers

[kpoltorak]

Homework Statement

Show that $$\forall a,b \in R$$:
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

Homework Equations

Triangle Inequality seems to be useless.

The Attempt at a Solution

$$(a+b)^{2}=a^{2}+b^{2}+2ab$$
$$2ab=(a+b)^{2}-(a^{2}+b^{2})$$
$$ab=\frac{1}{2}(a+b)^{2}-\frac{1}{2}(a^{2}+b^{2})$$
$$\left|ab\right|=\left|\frac{1}{2}(a+b)^{2}-\frac{1}{2}(a^{2}+b^{2})\right|$$
$$\left|ab\right|=\left|\frac{1}{2}(a^{2}+b^{2})-\frac{1}{2}(a+b)^{2}\right|$$

 

[Mentallic]

Well you pretty much have it. Take a look at the right hand side, they all consist of squares, so...?

 

[kpoltorak]

Right I've noticed that the values on both sides of the minus sign are all positive, however that doesn't necessarily mean that $$\left|ab\right|$$ is less than $$\frac{1}{2}(a^{2}+b^{2})$$. Because its an absolute value, the LHS of the minus sign could be smaller than the RHS while preserving the equality. For example, it is NOT TRUE that $$\left|6\right|=\left|2-8\right|\rightarrow 6\leq 2$$

 

[Mentallic]

Well yes I realize that, but you seem to have forgotten what the question is. Take a look at it again!
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

Notice how the right side only has that one term in there, and by the way, $$\frac{1}{2}(a+b)^2\geq 0$$

 

[kpoltorak]

I think I see...?

$$|ab|\leq\left|\frac{1}{2}(a^{2}+b^{2})-\frac{1}{2}(a+b)^{2}+\frac{1}{2}(a+b)^{2}\right|$$
$$\left|ab\right|\leq\left|\frac{1}{2}(a^{2}+b^{2})\right|$$
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

 

[Mentallic]

You shouldn't have the term you added inside the absolute value sign. For example,

$$|10|=|5-15|$$

$$|10|\leq |5-15+15|=|5|$$ is obviously wrong.

Instead, you should consider both cases when the RHS inside the absolute value is more than zero, and then less than zero and show both cases hold true for the inequality you want to prove.

 

[kpoltorak]

Now I see. Thank you for all your help!

 

[Mentallic]

You're welcome

 

[Tom McCurdy]

Here was the way I thought about it:
$$|ab| \leq \frac{1}{2}(a^2+b^2)$$
times by 2 and square both sides

$$4a^2b^2 \leq (a^2+b^2)^2$$

$$0 \leq a^4 - 2a^2b^2 + b^4$$

$$0 \leq (a^2-b^2)^2$$

and since it is squared it must be greater than or equal to zero

 

[Mentallic]

That's much more elegant Tom.