MHB Proving absolute values theorems

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To prove the theorem that if \( x + y \ge 0 \) then \( |x + y| = x + y \), we start by defining \( a = x + y \). The absolute value of \( a \) is defined as \( |a| = a \) when \( a \ge 0 \). Given that \( a \) is non-negative, we directly conclude that \( |x + y| = x + y \). This proof relies solely on the definition of absolute value and the initial condition provided. Therefore, the theorem is validated.
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For all real numbers $x$ and $y$ , if $x + y >= 0$ then $|x + y| = x + y$. How would I prove this?

My textbook just assumes this to be true.
 
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Let $a = x+y$. What is $|a|$?
 
Deveno said:
Let $a = x+y$. What is $|a|$?

$|x + y|$
 
tmt said:
For all real numbers $x$ and $y$ , if $x + y >= 0$ then $|x + y| = x + y$. How would I prove this?

My textbook just assumes this to be true.

The absolute value of a real number is defined as:
$$|a| = \begin{cases}a&\text{if } a\ge 0 \\ -a & \text{if } a<0\end{cases}$$

Since it is already given that $a = x+y \ge 0$, it follows from the definition that $|x+y|=x+y$.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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