Is this proof of with absolute values and inequalities correct?

Question

Show,

$||a − c| − |b − d|| ≤ |a − b| + |c − d||$

To show this, we use the property of absolute values,

$|x|\le y$ iff $-y\le x\le y$

Applying this to our inequality gives,

$|b − a| + |d − c|≤|a − c| − |b − d| ≤ |a − b| + |c − d||$

For, $-y\le x$

Using the Absolute Value-Square Root Theorem,

$\sqrt{(b − a)^2} + \sqrt{(d − c)^2}≤\sqrt{(a − c)^2}− \sqrt{(b − d)^2}$

Squaring both sides,

$ {a^2}+{b^2}+{c^2}+{d^2}-2ab-2cd+2bd-2bc-2ad+2ac ≤{a^2}+{b^2}+{c^2}+{d^2}-2ab+2cd+2bd+2bc-2ad-2ac$

Simplifying leaves us,

$ -(d-a+b)≤(d-a+b)$

So, we have shown $-y\le x$ for any a,b,c and d.

For $x\le y$. We do the same thing: use the above mentioned theorem, square and simplify. However, this leaves $0=0$. Does this prove $x\le y$ ?

I'm still learning proofs so I apologize if this is elementary stuff. Is this enough to prove my statement? It seems like it might not communicate what I want to show although I'm fairly sure I'm close

Answer 1

There are several errors in your proof:

• Squaring both sides of an inequality is produces an equivalent inequality only if both sides are known to be non-negative. That is not the case here, the right-hand side $\sqrt{(a − c)^2}− \sqrt{(b − d)^2}$ can be negative.

• Squaring both sides of the inequality is done incorrectly.

• Your final inequality $-(d-a+b) \le (d-a+b)$ need not be true. So even if all previous steps were correct (which they are not), this would not prove the desired estimate.

Note that it much is easier to use the triangle inequalities $||x|-|y|| \le |x \pm y| \le |x| + |y|$:

$$ ||a − c| − |b − d|| \le |(a-c) - (b-d)| = |(a-b) - (c-d)| \le |a − b| + |c − d| \, . $$

Answer 2

Your idea is right.

After squaring of the both sides we need to prove that: $$|(a-c)(b-d)|+|(a-b)(c-d)|\geq ab+cd-ac-bd$$ or $$|ab+cd-ad-bc|+|ad+bc-ac-bd|\geq ab+cd-ac-bd,$$ which is true by the triangle inequality: $$|ab+cd-ad-bc|+|ad+bc-ac-bd|\geq |ab+cd-ac-bd|\geq ab+cd-ac-bd.$$