How do I prove this? x and y are real numbers. Thanks for the help.
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4Have you tried squaring both sides? – Gautam Shenoy May 19 '16 at 06:20
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1See also: http://math.stackexchange.com/questions/688869/how-do-i-prove-sqrtx2-y2-le-x-y – colormegone May 19 '16 at 06:33
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1Think of $|x|$ and $|y|$ as sides of a right angled triangle. – Aritra Das May 19 '16 at 06:37
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Since both sides of $\sqrt{x^{2} + y^{2}} \le |x| + |y|$ are non-negative real numbers, taking the square on both sides gives the equivalent statement $$x^{2} + y^{2} \le (|x| + |y|)^2 = |x|^2 + |y|^2 + 2|x||y| = x^2 + y^2 + 2|xy| \tag{1}\label{*} $$
Since we define the absolute value on the reals as $|x|= \sqrt{x^2}$
So then clearly ${|x|}^2=x^2$
Also we know $2|xy| \geq 0$ for all real numbers $x,y$, and $(1)$ follows.
Walt van Amstel
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lattice
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This is the same answer I arrived at but, I wasn't to sure whether x^2 + y^2 was the same as |x|^2 + |y|^2 for all real numbers x and y since we could be dealing with negatives as well. Thank you. – EtherealMist May 19 '16 at 06:32
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Consider a right triangle with legs of length $|x|$ and $|y|$. Then the left side is the length of the hypotenuse, which by the triangle inequality is necessairly shorter than the sum of lengths of two other sides.
I hope this helps $\ddot\smile$
dtldarek
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Let's try proof by contradiction. Suppose instead there exist numbers $x$ and $y$ such that $\sqrt{x^2 + y^2} > |x| + |y|$. We then have that $x^2 + y^2 > x^2 + y^2 + 2|x|\cdot |y|$, which implies that $|x|\cdot |y| < 0$. This impossible (thnk about why). Therefore our hypothesis must be false and $\sqrt{x^2 + y^2} \leq |x| + |y|$ for all real numbers $x,y$.
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Thanks for the response. This was an interesting take on the proof. – EtherealMist May 19 '16 at 06:35