Prove $\forall m\in\mathbb{N},m\neq1:\quad\sum_{n=1}^{m}\frac{1}{n^2}\leq\int_1^m\frac{\sqrt{x^6+4}}{x^3}\ dx$

Question

So I have stumbled upon this question and was very intrigued on how to solve it. I have an intuitive solution, but I guess that's not enough. I would be glad if you could shed some light on how to make it more rigorous.

Prove the following inequality:

$$\forall m\in\mathbb{N},m\neq1:\quad\sum_{n=1}^{m}\frac{1}{n^2}\leq\int_1^m\frac{\sqrt{x^6+4}}{x^3}\ dx$$

My intuitive solution:

Define $f(x)=\frac{1}{x^2}$ on the interval $[1,m]$. Intuitively, the summation of the values of $f$ along its graph on the interval $[1,m]$ would be greater than the discrete summation of the series, since every element of the series is already being summed in the integral; alongside with other positive (real) values.

Thus, defining:

$$\gamma(t)=\left(t,\frac{1}{t^2}\right) \implies \gamma'(t)=\left(1,-\frac{2}{t^3}\right) \\ t\in[1,m]$$

We can say that:

$$L(\gamma)\equiv\int_\gamma ds\geq\sum_{n=1}^m\frac{1}{n^2}$$

Evaluating $\displaystyle \int_{\gamma}ds$, we will get the desired inequality:

$$\fbox{$\int_1^m \frac{\sqrt{t^4+6}}{t^3}\ dt \geq \sum_{n=1}^{m}\frac{1}{n^2}$}$$

I would be glad to hear your thoughts. Thanks!

P.S.: Since the integrand is greater than $1$, we can say that when $m\to\infty$, the integral diverges. Thus, since the series converges, the inequality becomes trivial for large enough values of $m$. The interesting values of the inequality would be as low as possible, specifically $m=2$.

Answer 1

We have $$\int_1^m\frac{\sqrt{x^6+4}}{x^3}\,\mathrm dx\ge \int_1^m\frac{\sqrt{x^6}}{x^3}\,\mathrm dx =\int_1^m\,\mathrm dx=m-1$$ whereas $$\sum_{n=1}^m\frac1{n^2}\le 1+\sum_{n=2}^m\frac14=1+\frac{m-1}4. $$ This solves $m\ge 3$, we need only check if $$ \int_1^2\frac{\sqrt{x^6+4}}{x^3}\,\mathrm dx\ge\frac54?$$

Answer 2

@Hagen von Eitzen shows that it holds for $m\ge 3$. Still it need to check if $$\int_1^2\frac{\sqrt{x^6+4}}{x^3}\,\mathrm dx\ge\frac54.$$ This answer will prove that it holds for $m=2$. Let $f(x)=\frac{\sqrt{x^6+4}}{x^3}=\sqrt{1+\frac{4}{x^6}}$.

Assume $g(x)$ is a function on $[1,2]$ such that $f(x)> g(x)$ and $\int_1^2 g(x)\mathrm dx \ge \frac{5}{4}$. Then we have $$\int_1^2 f(x)\mathrm dx > \int_1^2 g(x) \mathrm dx \ge \frac{5}{4}.$$

To find such function $g(x)$, I use Octave code (https://octave-online.net/bucket~HZvybAz3FzZBX2cvzgWg86 ) to find a class of function $g(x)=1+\frac{a_1}{x^{b1}}-\frac{a2}{x^{b2}}$. One solution is $(a_1,a_2,b_1,b_2)=(2,1,3,2)$. Thus $g(x)=1+\frac{2}{x^3}-\frac{1}{x^2}$ should be an option. However it still needs to prove that $g$ do satisfy the conditions.

• It follows from $1\le x \le 2$ that $x>0$ and $x^3-x+2>0$. Thus we have $$\begin{split} &f(x)>g(x) \\ \Longleftrightarrow &\sqrt{1+\frac{4}{x^6}}>1+\frac{2}{x^3}-\frac{1}{x^2} \\ \Longleftrightarrow &\sqrt{x^6+4}>x^3-x+2 \\ \Longleftrightarrow &x^6+4>(x^3-x+2)^2 \\ \Longleftrightarrow &x (2 x^3 - 4 x^2 - x + 4)>0 \\ \Longleftrightarrow &2 x^3 - 4 x^2 - x + 4>0 \\ \end{split}$$ It is easy to calculate that $\min_{1\le x\le 2}{2 x^3 - 4 x^2 - x + 4}=\frac{58}{27}>0$. So $f(x)>g(x)$.
• $\int_1^2 g(x)\mathrm dx =\int_1^2 \left(1+\frac{2}{x^3}-\frac{1}{x^2}\right) \mathrm dx = \frac{x^3+x-1}{x^2}\Big|_1^2= \frac{5}{4}$

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As a complete proof, I would like to use another method, "induction on $m$", to prove the original question.

For $m=2$, it is proved in the above. Assume that it holds for $m\ge 3$, i.e., $\sum_{n=1}^{m}\frac{1}{n^2}\leq\int_1^m\frac{\sqrt{x^6+4}}{x^3}\mathrm dx$.

Considering that $$\int_1^{m+1}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx=\int_1^m\frac{\sqrt{x^6+4}}{x^3}\mathrm dx+\int_{m}^{m+1}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx$$ and $$\sum_{n=1}^{m+1}\frac{1}{n^2}=\left(\sum_{n=1}^{m}\frac{1}{n^2}\right)+\frac{1}{({m+1})^2},$$ by induction it is sufficient to prove $$\int_{m}^{m+1}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx\ge \frac{1}{({m+1})^2}.$$ Obviously, it is true since when $m\le x \le m+1$, $$\frac{\sqrt{x^6+4}}{x^3}\ge \frac{\sqrt{(m+1)^6+4}}{(m+1)^3}=\sqrt{1+\frac{4}{(m+1)^6}}> 1 \ge \frac{1}{({m+1})^2}.$$ So $$\int_{m}^{m+1}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx>\int_m^{m+1} \frac{1}{({m+1})^2} \mathrm dx = \frac{1}{({m+1})^2}.$$

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Remark: As the "P.S" added at the end of the OP, the author notices that $A=\int_{n}^{n+1}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx$ is much greater than $B=\frac{1}{({n+1})^2}$ when $n$ is large enough. In fact $A-B\to 1$ when $n\to +\infty$ and it is obvious that $A>1>B$. If we rewrite $\int_1^m\frac{\sqrt{x^6+4}}{x^3}\mathrm dx$ as $S_1=\sum_{n=2}^{m}\int_{n-1}^{n}\frac{\sqrt{x^6+4}}{x^3}\mathrm dx$ and compare it with $S_2=\sum_{n=1}^{m}\frac{1}{n^2}$, then you will find that the $n$-th item in $S_1$ is always greater than the $n$-th item in $S_2$ except $n=2$. However note that the number of items in $S_1$ is exact one less than that in $S_2$. So the first one item in $S_1$ has to corresponding to first two items in $S_2$. So I agree to Amit Zach's comment that it is more difficult to prove it for $m=2$.