# given positive real numbers x,y how to demonstrate that is defined as the maximum of x and y?

Is it true that given positive real numbers $x,y$, then we have that

$$\sqrt{x^2 + y^2} \geq \max\{ x, y \}$$

I cant find a counter-example although it seems it is true... Any comments?

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This is indeed true and follows directly from the fact that $y^2,x^2\ge0$ hence $|x|\sqrt{1+(y/x)^2}\ge |x|$ (conversely the same is true for $y$) hence the result – b00n heT Feb 2 '14 at 10:57
why didn't you accept my answer? (I was the first to post...) – etothepitimesi Feb 2 '14 at 11:19
I did it now! thank you very much for your help! – ILoveMath Feb 2 '14 at 12:22

We know that

$${x^2} + \underbrace {{y^2}}_{ \ge 0} \ge {x^2}$$

$$\underbrace {{x^2}}_{ \ge 0} + {y^2} \ge {y^2}$$

So, taking the square root on both sides of each expression, we get

$$\sqrt {{x^2} + {y^2}} \ge \left| x \right| \ge x$$

$$\sqrt {{x^2} + {y^2}} \ge \left| y \right| \ge y$$

Thus

$$\sqrt {{x^2} + {y^2}} \ge \max \left\{ {x,y} \right\}.$$

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Yes. To see this, note that $$\tag{1}\sqrt{x^2+y^2}\geq \sqrt{x^2}=x$$ since $x$ is positive. Similarly, we have $$\tag{2}\sqrt{x^2+y^2}\geq \sqrt{y^2}=y$$ since $y$ is positive. Combining $(1)$ and $(2)$, we have $$\sqrt{x^2+y^2}\geq\max\{x,y\}.$$

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