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I need to prove the following result:

Let $x, y \in\mathbb{R}$. Prove that $|x| + |y| \geq |x+y|$

I know this is the triangle inequality, but I haven't seen one version that helps me solve this one. Can someone help me work this step by step?

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    $\begingroup$ you may want to reverse that inequality sign ... $\endgroup$ – hickslebummbumm Oct 17 '14 at 0:31
  • $\begingroup$ What set are x and y in? It looks like it got cut off in your problem statement. I also second the comment above mine. The way you have it written is not going to be true in general. $\endgroup$ – epsilonics Oct 17 '14 at 0:31
  • $\begingroup$ You're right that i want to reverse it. and no, the question did not get cut off. $\endgroup$ – yyjd Oct 17 '14 at 0:34
  • $\begingroup$ I did include the whole question. That's all I have to go off of. The only other thing included is (this is called the triangle inequality). $\endgroup$ – yyjd Oct 17 '14 at 0:37
  • $\begingroup$ Please, Bye_World. I know this is easy for you but to me this doesn't make any sense. Can you help walk me through this? Please? $\endgroup$ – yyjd Oct 17 '14 at 0:42
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A proof

Usually is done by taking the square of both sides as:

$$(|x|+|y|)^2 \ge (|x + y|)^2 $$

If you develop this expression

$$(|x|+|y|)^2 = |x|^2 + 2|x||y| + |y|^2 $$

and

$$(|x + y|)^2 = (x+y)^2 = x^2 + 2xy + y^2 $$

as $|\cdot|^2$ is the same as just $(\cdot)^2$

Now, if you compare the two expressions:

$$ |x|^2 = x^2 $$ $$ |y|^2 = y^2 $$

But

$$2|x||y| \ne 2xy$$

Because $x$ and $y$ can have different sign. That is why,

$$2|x||y| \ge 2xy$$

Therefore

$$(|x|+|y|)^2 \ge (|x + y|)^2 $$

If you generalize this using the norm $||\cdot||$ this inequality says that the sum of of two vectors will be maximum only when they have the same direction and orientation

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The way to prove this will vary depending on what set $x$ and $y$ are elements of. I am typing this on my phone, so in case you did put down the set they are in and I can't see it because of my phone browser, apologies. I'll just assume they are in the reals.

You have three cases, $x$ and $y$ both positive, $x$ and $y$ both negative, and $x$ negative, $y$ positive.

In the first case, $|x+y| = |x|+|y|$.

The second case is the same, for example $|-4 + -7| = |-11| = 11$, and $|-4|+|-7| = 4+7 = 11$.

The last case is where the inequality comes from. Let $ x $ be negative and $ y $ be positive. $x < |x|$ and $ y=|y|$, so $ x + y < |x| + y$, and then take the absolute value of both sides to get $|x+y|< ||x|+y| = |x|+|y|$.

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