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?

  • 3
    $\begingroup$ 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

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

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


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

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

Now, if you compare the two expressions:

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


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

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

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


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

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


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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