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Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?

How do I deal with square roots inside the inequality?

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  • $\begingroup$ Thank you, Michael. I'm leaving the accepted answer open for now as I am also looking for a pre-calculus solution. $\endgroup$ – user17982 Aug 14 '13 at 7:46
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    $\begingroup$ Fair enough. In future you should make such requests in the question. $\endgroup$ – Michael Albanese Aug 14 '13 at 7:55
  • $\begingroup$ Isn't this fairly simple? Take vectors $(x,1)$ and $(y,1)$, and apply the triangle inequality. (Or is that how you came up with this?) $\endgroup$ – Chris Sep 8 '13 at 19:19
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If you are familiar with Minkowski's inequality, this is straightforward.

$$\sqrt{x^2+1}+\sqrt{y^2+1} \ge \sqrt{(x+y)^2 + (1+1)^2} = \sqrt5$$

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  • $\begingroup$ Really, this is just the triangle inequality for vectors $(x,1)$ and $(y,1)$. (+1, however) $\endgroup$ – Chris Sep 8 '13 at 19:17
  • $\begingroup$ With the added comment of user1296727...this is the best answer present here...I really like this... $\endgroup$ – Hawk Apr 27 '14 at 16:34
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Triangle inequality :) The length of the kinked red line represents the sum of the two roots, which is not shorter that the long diagonal of this $1:2$ rectangle.

enter image description here

(Edit: changed my poorly drawn diagram; was this)

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    $\begingroup$ Very nice. I was just writing up an answer similar to this. $\endgroup$ – Michael Albanese Aug 14 '13 at 8:04
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    $\begingroup$ Very nice! If anyone is struggling to see why this is proof, think what the shortest length possible for the kinked red line is. The black line across the middle of the picture is split into two lengths x and y which sum to 1. Where this division occurs is where the red lines meet. The shortest that the sum of the lengths these lines can be is thus the diagonal, sqrt 5! $\endgroup$ – JH92 Aug 14 '13 at 8:33
  • $\begingroup$ @peterwhy great answer! $\endgroup$ – Sepideh Bakhoda Aug 14 '13 at 9:14
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Assuming the contrary, for some $0\le x \le 1$, and $y=1-x$,

$$\begin{align} \sqrt{x^2+1}+\sqrt{y^2+1} <& \sqrt{5}\\ \left( \sqrt{x^2+1} + \sqrt{y^2+1} \right)^2 <& 5 &\text{(Square both sides)}\\ x^2 +1 + 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} + y^2 + 1 <& 5\\ 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - x^2 -y^2\\ 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - \left(x+y\right)^2 + 2xy\\ 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - 1 + 2xy &\text{(Substitute }x+y=1\text{)}\\ \sqrt{\left( x^2+1 \right) \left( y^2 + 1 \right)} <& 1+xy\\ \left( x^2+1 \right) \left( y^2 + 1 \right) <& \left( 1+xy \right)^2 &\text{(Square both sides)}\\ x^2 y^2 + x^2 + y^2 + 1 <& 1 + 2xy + x^2 y^2\\ x^2 -2xy+y^2 <& 0\\ \left( x - y \right)^2 <& 0\\ \end{align}$$

which contradicts.

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  • $\begingroup$ It'd be nice if you'd explain what happened from line 5 to line 6 (the step where you wrote "both sides are positive") $\endgroup$ – DonAntonio Aug 14 '13 at 9:22
  • $\begingroup$ Substituted $x+y=1$, squaring both sides, removing common 2 on both sides $\endgroup$ – peterwhy Aug 14 '13 at 9:26
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    $\begingroup$ Thanks @peterwhy: now we all know. Very nice solution +1 $\endgroup$ – DonAntonio Aug 14 '13 at 9:31
  • $\begingroup$ I would be interested to see how to write this proof in a contradiction-free way. $\endgroup$ – user17982 Aug 14 '13 at 14:14
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    $\begingroup$ Write from the bottom of my proof up to your result, and change all $<$ to $\ge$ (i.e. negate all statements) $\endgroup$ – peterwhy Aug 14 '13 at 14:17
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Let $f(x) = \sqrt{x^2+1}+\sqrt{(1-x)^2+1}$. As $x$ and $y$ are non-negative integers, we only consider $x \in [0, 1]$.

Note that $f(x)$ is symmetric about $x = \frac{1}{2}$ and

$$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}}.$$

For $0 < x < \frac{1}{2}$, $x < 1 - x$ so $\sqrt{x^2+1} > \sqrt{(1-x)^2+1}$. Therefore, $$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < \frac{\frac{1}{2}\sqrt{(1-x)^2+1} - \frac{1}{2}\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < 0.$$ So $f$ is decreasing on $[0, \frac{1}{2})$ and as $f$ is symmetric about $x = \frac{1}{2}$, $f$ is increasing on $(\frac{1}{2}, 1]$. Therefore $f$ attains its minimum value at $x = \frac{1}{2}$ which is $$f\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^2+1} + \sqrt{\left(1-\frac{1}{2}\right)^2+1} = \sqrt{\frac{5}{4}} + \sqrt{\frac{5}{4}} = \frac{1}{2}\sqrt{5} + \frac{1}{2}\sqrt{5} = \sqrt{5}.$$ Hence, $f(x) \geq \sqrt{5}$ for $x \in [0, 1]$. Setting $y = 1 -x$ we have $\sqrt{x^2+1} + \sqrt{y^2+1} \geq \sqrt{5}$.

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Another interpretation of the problem is this: $ \ y = \sqrt{x^2 + 1} \ \ \text{and} \ \ x = \sqrt{y^2 + 1} \ $ are the "positive" branches of the "vertical" and "horizontal" unit hyperbolas centered on the origin. They thus share the asymptotes $ \ y = \pm \ x \ $ .

It is specified that $ \ x \ge 0 \ \ \text{and} \ \ y \ge 0 \ $ and that they be related through $ \ x + y = 1 \ . $ So we in fact require that $ \ 0 \ \le \ x,y \ \le \ 1 \ . $ We can thus imagine a point in the first quadrant which slides along the line $ \ y = 1 - x \ $ in said quadrant and consider the lengths of the vertical and horizontal line segments extending from each coordinate axis to the appropriate hyperbolic branch, as depicted in the figure below.

enter image description here

It is clear that the behavior of this geometric arrangment is symmetrical about the asymptote $ \ y = x \ , $ so we only need look at the results for, say, $ \ 0 \ \le \ x \ \le \ \frac{1}{2} \ . $ For $ \ x = 0 \ , $ we attain the maximum for the sum sought,

$$ \sqrt{0^2 + 1} \ + \ \sqrt{1^2 + 1} \ = \ 1 + \sqrt{2} \ \approx \ 2.4142 \ , $$

while for $ \ x = \frac{1}{2} \ , $ the sum has its minimum,

$$ \sqrt{(\frac{1}{2})^2 + 1} \ + \ \sqrt{(\frac{1}{2})^2 + 1} \ = \ 2 \cdot \sqrt{\frac{5}{4}} \ = \ \sqrt{5} \ \approx \ 2.2361 \ . $$

The sum of radicals has a fairly narrow range of values for the specified domain. [This approach is most nearly similar to Michael Albanese's analysis. (It also seems a sort of "inside-out" version of peterwhy's elegant method.)]

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Construct a right-angled triangle, $1$ unit and $2$ units long for two perpendicular sides. The length of hypotenuse is? And note that straight line gives the shortest distance between two points.

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