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I have to show that the following inequality is true $$ \frac{A}{\sqrt{n - 1}} + \frac{4}{n} < 2 \frac{A}{\sqrt{n}} $$ for $n \in \mathbb{N} \backslash \{1\}$ and $A \in \mathbb{R}_{+}$. Ideally, I would like to prove it for any $A \in \mathbb{R}_+$, but I would settle for a lower bound on $A$ as long as I can prove the statement for any $n \in \mathbb{N}$.

I tried to prove this statement by induction on $n$ and I managed to show that it holds true for $n = 2$ under the condition that $A > 4.82...$. What I am having problems with is the inductive step. Any help with it or suggestion about a different approach to tackle the problem would be welcome.

Thanks.

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  • $\begingroup$ We should prove this inequality without any conditions on $A$? $\endgroup$ – Dr. Sonnhard Graubner May 7 '18 at 16:37
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We have $$ \frac 2{\sqrt n}-\frac1{\sqrt {n-1}}=\frac{2\sqrt{n-1}-\sqrt n}{\sqrt n\sqrt{n-1}}=\frac{4(n-1)-n}{\sqrt n\sqrt{n-1}(2\sqrt{n-1}+\sqrt n)}>\frac{3n-4}{\sqrt n\sqrt n\cdot 3\sqrt n}=\frac{3n-4}{3n\sqrt n}$$ and we want $A$ times the left hand side to be $>\frac 4n$. For this, it suffices to have $$ \frac{3n-4}{3n\sqrt n}A>\frac{4}{n}$$ or,$$A>\frac{12\sqrt n}{3n-4}.$$ For $n\ge 3$, the right hand side is $ \le\frac{12\sqrt n}{\frac 53n}=\frac{36}{5\sqrt n}\le \frac{36}{5\sqrt 3}<\frac92.$ On the other hand, the exact conditions on $A$ for $n=2$ demand $$A>\frac 2{\sqrt 2-1} =2\sqrt 2+2.$$

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  • $\begingroup$ Thank you very much. This solves the problem! $\endgroup$ – F.Vitiello May 7 '18 at 17:25
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The inequality is equivalent to $$2\sqrt{n}-\frac n{\sqrt{n-1}}>\frac4A.$$ The left-hand side is a strictly increasing function of $n$ that converges to $+\infty$ as $n$ approaches infinity. Hence, the inequality holds for all $n\geq2$ as long as it holds for $n=2$. This means $$2\sqrt{2}-2>\frac 4A$$ or, equivalently, $$A>\frac2{\sqrt{2}-1}.$$

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  • $\begingroup$ Clearly the better answer. $\endgroup$ – trancelocation May 7 '18 at 18:28
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For example we get if $n=5$:$$\frac{4}{5}<A\left(\frac{2\sqrt{5}}{5}-\frac{1}{2}\right)$$ so $$\frac{8}{4\sqrt{5}-5}<A$$

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  • $\begingroup$ Well, this is true but what I am looking for is a proof of the statement for all $n \in \mathbb{N}$. This is the reason why I tried using induction, it seemed the most natural way to prove this. Thanks anyway. $\endgroup$ – F.Vitiello May 7 '18 at 16:56

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