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Is it valid to say that since $\sqrt{n(n + 1)} > 1$ and $n + \frac{1}{2} > 1$ that we can square both sides without having to assume anything about their inequality?

Also can someone show me how with induction or strong induction?

joriki
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mk3009hppw
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  • AM-GM: $$\sqrt{n\cdot(n+1)}\leq\frac{n+(n+1)}2=n+\frac12$$ – Rushabh Mehta Dec 18 '19 at 21:22
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    Is it not valid to square both sides? – mk3009hppw Dec 18 '19 at 21:23
  • It's not that it is invalid but it is a slightly more intrinsic argument ($x^2$ needs to be a strictly increasing function over the naturals). AM-GM is simply more general and useful. – Rushabh Mehta Dec 18 '19 at 21:25
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    You indeed can square both sides, but because they're both nonnegative, not because they're greater than $1$. – Bernard Dec 18 '19 at 21:35
  • Yes but if one is greater than 1 at some point and one is less than 1, then one will decrease, one will increase. But if they both are greater than 1 always, then the functions are clearly monotonically increasing. Anyway, thanks for the help. – mk3009hppw Dec 18 '19 at 21:37
  • Monotonically increasing means $f(x) \leq f(y)$ whenever $x \leq y$. You are describing $f(x) \geq x$, which is sometimes a useful relationship, but not so much in this context. All that matters here is that $f(x) = x^2$ is monotonically increasing on $x \geq 0.$ – Brian Moehring Dec 18 '19 at 21:53

2 Answers2

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For all $n\in\mathbb{N}$, $$\sqrt{n(n+1)}<\sqrt{n^2+n+\frac{1}{4}}=\sqrt{\left(n+\frac{1}{2}\right)^2}=n+\frac{1}{2}$$

A. Goodier
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  • Man I am so dumb. How do people get so sharp at math. My prof went over 200 induction proofs like this and then put a simple algebra problem on the exam... – mk3009hppw Dec 18 '19 at 21:27
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    It's literally the same thing as squaring both sides though @mk3009hppw – Simply Beautiful Art Dec 18 '19 at 21:34
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    @mk3009hppw: This is the way is works for me at least. Sometimes you remember a trick sometimes you don't. Most things are blindingly obvious in retrospect. – copper.hat Dec 18 '19 at 21:53
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Squaring both sides you get $$ n^2+n\le n^2+n+\frac {1}{4}$$ which is true for all n.

Mohammad Riazi-Kermani
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