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I'm working out inequalities and trying to understand proofs with them. I'm currently trying to show that for $a \in \mathbb{N}$. I'm trying to prove with contradiction. I'm trying to show $\sqrt{a} \leq \sqrt{a+\sqrt{a}}$. So I'm supposing $\sqrt{a} > \sqrt{a+\sqrt{a}}$. Then we can multiply both sides by $\sqrt{a}$ to see $a > \sqrt{a+\sqrt{a}}\sqrt{a}$. We can also multiply by $\sqrt{a+\sqrt{a}}$ to see $\sqrt{a+\sqrt{a}}\sqrt{a} > a+\sqrt{a}$. Then from the order, we see that $a > a + \sqrt{a}$. Which means that $0 > \sqrt{a}$, which may be true if $\sqrt{a}=(-a)^2$, so I feel like there is no contradiction since there is a case where $0 > \sqrt{a}$. Am I missing something?

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    No, you aren't missing anything – 2017-02-25
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    What about squaring both sides? If $a,b\geq 0$, $a\geq b \Leftrightarrow a^2\geq b^2$. But also, are you really trying to prove that $\sqrt{a} \geq \sqrt{a+\sqrt{a}}$, not the *other* direction? (As a sanity check, take $a=1$. Do you *really* have $1\geq \sqrt{2}$?) – 2017-02-25
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    Awe shoot, I was trying to prove the other direction. I'm editing. – 2017-02-25
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    What's the point of proving by contradiction something that's obvious the direct way? – 2017-02-25

2 Answers 2

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The inequality is false unless $a =0$. For any $a > 0$ we have that $\sqrt{a} > 0$, and so $a < a + \sqrt{a}$. Taking the square root of both sides then shows that $\sqrt{a} < \sqrt{ a + \sqrt{a}}$.

EDIT: Now that you have flipped the inequality in the post, the above is actually a proof of the inequality you want.

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When dealing with inequalities which are positive on both sides, you can do the following:

$\sqrt{a} \leq \sqrt{a+\sqrt{a}}$ $\Leftrightarrow a \leq a + \sqrt{a} \Leftrightarrow$ $\sqrt{a} \geq 0 \Leftrightarrow a \geq 0$ which is true since $a \in \mathbb{N}$

Therefore the initial statement is true.

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    Why is it a contradiction? We can have $\sqrt{9} = -3$. Can't $\sqrt{x}$ map from $\mathbb{N} → \mathbb{R}$? – 2017-02-25
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    The question has been changed since I posted, so the statement is now true, but what I did was fine since we are given that $a \geq 0$ – 2017-02-25
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    It's never true that $\sqrt{9}=-3$. That's not how the square root symbol works. It's defined as the positive square root. – 2017-02-25
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    Yeah, I realized I made a mistake in the question, but I think your solution still is helping. My only concern is still what i commented before: Am I wrong in assuming $\sqrt{x}$ can map from $\mathbb{N} → \mathbb{R}$, so we can have $\sqrt{9} = -3$? – 2017-02-25
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    Yes you are wrong, $\sqrt{9} = 3$, this is how it is defined. It does not matter that $(-3)^2 = 9$, this is why $\sqrt{x^2} = |x|$ and not $\sqrt{x^2} = x$ – 2017-02-25
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    What aspect of my proof do you disagree with? – 2017-02-25