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Only a hint on how to prove this, if not a complete proof, would also be appreciated.

Noor Aslam
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  • Hi. Have you tried to answer this yourself? If so, could you give some methods you have tried in the question, so people can help you better. – Joe Tait Feb 24 '14 at 19:18
  • When confronted by annoying roots, square to get rid of them... – vonbrand Mar 07 '14 at 14:35

6 Answers6

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$$\sqrt{x^2+y^2}=\sqrt{|x|^2+|y|^2}\le \sqrt{|x|^2+|y|^2+2|x||y|}=\sqrt{(|x|+|y|)^2}=|x|+|y|$$

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    The OP did just ask for a hint ... – Thomas Feb 24 '14 at 18:58
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    @Thomas Well, but I did. Why did you inform me about your non-downvoting? – Michael Hoppe Feb 24 '14 at 19:15
  • @MichaelHoppe: I am sorry! I mistook your comment as you saying that I downvoted the answer. Sorry about that! (I of course would not downvote a competing answer) – Thomas Feb 24 '14 at 19:16
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    @Thomas "If not" means "perhaps even." So I do not see how a complete proof will hurt. – Lord Soth Feb 24 '14 at 19:23
  • @LordSoth: Let's not discuss that here... – Thomas Feb 24 '14 at 19:24
  • Didn't I understand the OP's question? He asked for a hint or for a complete answer! But if you suggest that a hint is more useful for him that's OK @Thomas –  Feb 24 '14 at 19:27
  • @SamiBenRomdhane: Yes, I think giving a hint is more helpful. Just my opinion. (And just to make it clear: I didn't downvote you) – Thomas Feb 24 '14 at 19:28
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Note that $$ x^2 + y^2 \leq (\lvert x\rvert + \lvert y\rvert)^2 $$ (You see this by expanding the right hand side.)

Thomas
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For a geometric intuition, think of $(x, y)$ as a point on the plane, then consider the Pythagorean theorem.

dani_s
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By the Pythagorean Theorem, let $c=\sqrt{x^2+y^2}$ equal the length of the hypotenuse of a right triangle with points $(0,0) (0,x) (0,y)$. One side is on the $x$-axis, and the other side is on the $y$-axis. Then the hypotenuse will never be greater than the sum of the two sides. There's slightly more if you want to be formal, but that's how I would set it up.

Brian J. Fink
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You might think of the Cartesian coordinates as polar coordinates and note that the trigonometric functions sine and cosine are bounded above and below.

1028
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Try to square both sides of the inequality. That should point you in the proper direction.

DrkVenom
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