关于皮斯库诺夫《微积分学教程》中反向三角不等式证明里条件|x|>|y|必要性的疑问

Hey there, I totally get where you're coming from—when your math knowledge is still elementary, even "basic" proof reasoning can feel uncertain. Let's unpack this reverse triangle inequality proof from Nikolai Piskunov's Differential and Integral Calculus together!

First, let's restate the proof as it appears in the book, along with the context you mentioned:

The absolute value of a difference is no less than the difference of the absolute values of the minuend and subtrahend.
$$|x-y|\ge |x|-|y| \text{, } |x|>|y|$$
Proof. Let $x-y=z$, then $x=y+z$ and from what has been proved (referring to the standard triangle inequality: $\forall {x,y}\in\mathbb{R}(|x+y|\leq |x|+|y|)$)
$$|x|=|y+z|\leq |y|+|z|=|y|+|x-y|$$
whence
$$|x|-|y|\le |x-y|$$
thus completing the proof.

Now, addressing your uncompleted question about the necessity of the $|x|>|y|$ condition:

• First off, the entire proof doesn't actually depend on $|x|>|y|$ at all. The derivation holds for all real numbers x and y. Let's test cases where $|x| \leq |y|$ to confirm:
  • If x=2, y=5: $|x-y|=3$, $|x|-|y|=2-5=-3$. Clearly $3 \ge -3$ is true.
  • If x=-1, y=-4: $|x-y|=3$, $|x|-|y|=1-4=-3$. Again, $3 \ge -3$ holds.
• So why does the book include $|x|>|y|$? Because this is the only scenario where the inequality gives a non-trivial, useful bound. When $|x| \leq |y|$, $|x|-|y|$ is either negative or zero, and since absolute values are always non-negative, $|x-y| \ge |x|-|y|$ is automatically true—it's just a "free" statement that doesn't tell us anything new about $|x-y|$.

The key takeaway here is: the reverse triangle inequality $|x-y| \ge |x| - |y|$ is valid for all real x and y. The $|x|>|y|$ condition is just highlighting the case where this inequality actually provides meaningful information (a positive lower bound for $|x-y|$), not a requirement for the proof to hold.

备注：内容来源于stack exchange，提问作者Dongchen