0
votes
2answers
24 views

Limit of |x-2| as x approaches -2

I believe that it equals -4. In the epsilon-delta definition, we can set delta equal epsilon and I become this satisfies the definition. The problem is I can't seem to prove based on this that 0 less ...
1
vote
4answers
128 views

Proving the inequality $|a-b| \leq |a-c| + |c-b|$ for real $a,b,c$

Let $a,b,c$ real numbers. Prove the inequality $|a-b| \leq |a-c| + |c-b|$. Prove that equality holds if and only if $a \leq c \leq b$ or $b \leq c \leq a$. I've tried starting with just $a \leq ...
0
votes
2answers
46 views

$|a-b|+|b-c|+|c-a|=2(\max\{a,b,c\}-\min\{a,b,c\})$

Let $a,b,c ∈ \Bbb R$ Show that $|a-b|+|b-c|+|c-a|=2(\max\{a,b,c\}-\min\{a,b,c\})$ Not sure where to start
1
vote
2answers
817 views

Sum of absolute values and the absolute value of the sum of these values?

I'm working on a proof and I need some help with this: I determined that for some situations ($x$ or $y$ are negative but not both): $|x| + |y| > x + y$ How can I conclude using that statement ...
5
votes
3answers
133 views

How would I prove $|x + y| \le |x| + |y|$?

How would I write a detailed structured proof for: for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$ I'm planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ ...
0
votes
1answer
58 views

how to prove if $a|b$ and $b\neq 0$, then $|a|\leq|b|$

where the conditions are: $a \neq 0$, $b \neq 0$ and $a$ and $b$ are integers. maybe i'm missing something very basic about the properties of an absolute values. My approach was to supposed, on the ...
4
votes
2answers
353 views

Spivak Calculus 3rd ed. $|a + b| \leq |a| + |b|$

I'm working through the first chapter of Michael Spivak's Calculus 3rd ed. Towards the end of the chapter he proves $ |a + b| ≤ |a| + |b| $ using the observation that $|a|= \sqrt{ a^2 }$ when $a$ ...
4
votes
3answers
491 views

Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?

Claim: If f is integrable, $\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx$ Proof (attempt): We know $-|f|\le f \le|f|$, so $\int-|f| \le \int f \le \int|f|$.* Since, if $-b<a<b$, we say ...