Is Proof method valid

Question

So I am studying for an upcoming midterm, and I am practicing my proofs. I found an old test online, that states the following:

$| x + y | + | x − y |≥| x | + | y |, x, y∈R .$

I want to know if my proof or attempt is fair enough to prove it?

my attempt:

let x be any arbitrary number, and since the absolute value of a number is always positive or zero

then $|x| ≥ x $ then $ |x+y| ≥ x+y ≥ |x| ≥ x $

(given that y is an element of the real number set) --->$|x+y|+|x| ≥ |x|+|y| +x ≥ |x|+|y| ≥ x+y ≥ x $

and lastly $ |x+y| +|x-y| ≥ |x|+|y|+|x| ≥ |x|+|y|≥ x+y ≥ x $

end proof.

It is not sound. This is not the only place where in incorrect assertion is made. For example, next line, it is not necessarily true that $|x|+|y|+x\ge |x|+|y|$, since $x$ could be negative. — André Nicolas, Oct 23 '15 at 06:12
I see, thank you, well that's why attempting random problems now is useful to learn — learnmore, Oct 23 '15 at 06:16
I wrote out a proof that does not use the Triangle Inequality. But the Triangle Inequality is very important, so it is better to use it. — André Nicolas, Oct 23 '15 at 06:38

score 1 · Answer 1 · answered Oct 23 '15 at 06:33

Note that if $s$ and $t$ are non-negative, then $s\ge t$ if and only if $s^2\ge t^2$. Now $$(|x+y|+|x-y|)^2=2x^2+2y^2+2|x^2-y^2|\ge 2x^2+2y^2\tag{1}$$ and $$(|x|+|y|)^2=x^2+y^2+2|xy|.\tag{2}$$ So it is enough to show that $2x^2+2y^2\ge x^2+y^2+2|xy|$, or equivalently that $x^2+y^2\ge 2|xy|$.

But this last inequality is clear, for $x^2+y^2-2|xy|=(|x|-|y|)^2$.

skyking · Answer 2 · 2015-10-23T06:49:04.417

No, for example the step $x+y\ge |x|$ is not true, the counter example would be if $x+y<0$. Then there's more errors further down, but it's hard to tell which are a consequence of the first error and which are new errors, but for example the step $|x|+|y|+x\ge|x|+|y|$ definitely looks odd and the step $x+y\ge x$.

Finally it's not nice to keep irrelevant parts in the proof. For example the last step $|x+y|+|x-y|\ge...\ge|x|+|y|\ge...$ everything to the right of $|x|+|y|\ge$ is irrelevant (you was only to proove that $|x+y|+|x-y|\ge|x|+|y|$ - never forget what you're about to prove).

What you could do if you want to prove it using brute force is to divide it into the cases where the $x$, $y$, $x-y$ and $x+y$ are each positive or negative - there 16 combinations, some of which are impossible (you probably see those). It's a tedious way, but given you don't find any other way it's a possibility.

An easier way would be to use the triangle inequality:

$$|x+y| + |x-y| \ge |x+y+x-y| = |2x|$$

and then use this to prove it if $|x|\ge|y|$. How you us it for the case $|x|\le|y|$ you maybe can figure out yourself?

A simpler way is to apply $$|x+y| + |x-y| \ge |x+y+x-y| = |2x|$$ for $y$ directly instead of $x$ . Then add the two inequalities and the result directly follows . — Nizar, Oct 23 '15 at 06:29

score 0 · Answer 3 · answered Oct 23 '15 at 06:39

0

we have for any $x$ and $y$ (using triangle inequality) $$|x+y| + |x-y| \ge |x+y+x-y| = |2x|$$ and $$|x+y| + |y-x| \ge |x+y-x+y| = |2y|$$ Adding up we get $$2(|x+y| + |x-y| )\ge |2x|+ |2y| $$ hence $$|x+y| + |x-y| \ge |x|+ |y| $$

answered Oct 23 '15 at 06:39

Nizar

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Is Proof method valid

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