# Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. [closed]

Could you please give me a hint on how to prove the inequality below for $$k \geq 3$$?

$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$$

## closed as off-topic by John Omielan, The Count, Daniele Tampieri, José Carlos Santos, ronnoJul 22 at 16:59

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• Just keep on bashing this inequality, you will get the answer!! – Anand Jul 21 at 13:57
• You may have noticed that the question received several close votes. Including some context for the question might make the question less likely to get closed (or more likely to get reopened). – Martin Sleziak Jul 22 at 8:53

$$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$ $$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$ We know that $$\sqrt{k(k+1)}\le\dfrac{k+(k+1)}{2}$$ (AM–GM inequality) and $$\dfrac{(2k+3)(k-1)}{2k^2}=\dfrac{2k^2+k-3}{2k^2}$$ is clearly greater than $$1$$ (because $$k\ge3$$). This proves the inequality.

• Thanks a lot for your input. – JohnK Jul 22 at 6:20
• You are welcome. – Piquito Jul 22 at 22:10

We have $$\left(1+\frac{2}{k}+\frac{3}{4k^2}\right)^2-\frac{k(k+1)}{(k-1)^2}=1/16\,{\frac {16\,{k}^{5}-24\,{k}^{4}-64\,{k}^{3}+{k}^{2}+30\,k+9}{{k} ^{4} \left( k-1 \right) ^{2}}} \geq 0$$ for $$k\geq 3$$

• Might be even easier to also multiply the quadratic $(k-1)^2$ away. You end up with an expression of only $k$'s resp. $1/k$'s. – Nikolaj-K Jul 21 at 13:47

If $$k \geq 3,$$ then...

\begin{align} \frac{\sqrt{k(k+1)} }{k-1} &=\frac{\sqrt{k^2+k}}{k-1} \\ &\lt \frac{\sqrt{k^2+k+\frac 1 4}}{k-1}= \frac{k+ \frac 1 2}{k-1}=1+\frac{3}{2k-2} \\ \end{align}

Have to compare $$\frac{3}{2k-2}$$ with $$\frac{8k+3}{4k^2}$$ when $$\ \ k \geq 3$$

\begin{align}\frac{3}{2k-2}&=\frac{12k^2}{4k^2\cdot (2k-2)} \\ \frac{8k+3}{4k^2}&=\frac{(8k+3)(2k-2)}{4k^2 \cdot (2k-2)}=\frac{16k^2-10k-6}{4k^2 \cdot (2k-2)} \\ &=\frac{12k^2+4(k-3)^2+14(k-3)}{4k^2 \cdot (2k-2)}\end{align}

therefore when $$\ \ k \geq 3$$ $$\frac{3}{2k-2} \leq \frac{8k+3}{4k^2}$$ And

\begin{align} \frac{\sqrt{k(k+1)} }{k-1} & \lt 1+\frac{3}{2k-2} \\ &=1+ \frac{12k^2}{4k^2\cdot (2k-2)}\\ &\lt 1+\frac{16k^2-10k-6}{4k^2 \cdot (k-1)} \\ &=1+\frac{8k+3}{4k^2}=1+\frac{2}{k}+\frac{3}{4k^2} \end{align}

• Only $k>0$ is not enough for this inequality.Take $k=1$ – Dr. Sonnhard Graubner Jul 21 at 13:50
• @Dr.SonnhardGraubner indeed, any thoughts after the edit? – AmateurMathPirate Jul 22 at 9:36

You have to solve this inequality system:

$$\left\{\begin{matrix} & \\k \neq 1 \: \land \: k\neq0 & \\k(k+1)\geqslant0 & \\1+\frac{2}{k}+\frac{3}{4k^2}\geq 0 & \\ \frac{\sqrt{k(k+1)}}{k-1} \geq 0 & \\\left (\frac{\sqrt{k(k+1)}}{k-1}\right )^2\leq \left ( 1+\frac{2}{k}+\frac{3}{4k^2} \right )^2 \end{matrix}\right.$$

In this way you wil get the solutions of the inequality.

• Did you want to have the square outside the final parenthesis? – J. W. Tanner Jul 21 at 15:42
• Yes, sorry. It's my error. – Matteo Jul 22 at 14:41