# Prove inequality with MVT

I want to prove the following inequality using the first mean value theorem for integrals: \begin{equation*}1\leq \int_0^1\frac{e^{\frac{x}{\sqrt{2}}}}{1+x^2}\, dx\leq \frac{1}{4-2\sqrt{2}}e^{\left (1-\frac{1}{\sqrt{2}}\right )}\approx 1,1440\end{equation*}



The first mean value theorem for integrals is the following:

Let $f$ be continuous on $[a, b]$. Then there exists $\xi \in (a, b)$ with \begin{equation*}\int_a^bf(x)\, dx=f(\xi)(b-a)\end{equation*}

Right? 

How can we apply this proposition to show the inequality? I got stuck right now. Could you give me a hint?

• Indeed, your first step is correct. Now note that as a consequence, $\int_{[a,b]} f \le (b-a) \max_{\xi \in [a,b]} f(\xi),$ and similarly with $\ge$ and $\min_{\xi \in [a,b]} f(\xi).$ This reduces the problem to computing these extrema (and indeed, that is where the expressions above come from). – stochasticboy321 Apr 6 '18 at 21:15
• You could minimize and maximize that function of $\xi\in [0,1]$, I suppose. I don't see why you're going to all this trouble. The original approach was correct, except for a miscalculation in the problem itself. – Ted Shifrin Apr 11 '18 at 20:04
• It seems to give a totally different set of bounds, and worse ones, at that. – Ted Shifrin Apr 11 '18 at 20:14
• I don't understand what you're doing. You have a function of $\xi$. You know the original integral is equal to that function evaluated at some particular $\xi\in [0,1]$. Therefore the integral is between the minimum and maximum values of that function of $\xi$. Just use usual techniques to find those. – Ted Shifrin Apr 11 '18 at 20:28
• Yes, that's what I got too. (Please factor things when possible rather than multiplying out all the constants.) As I said, worse bounds. – Ted Shifrin Apr 11 '18 at 21:25

The function $$f(x):=\frac{\exp(x/\sqrt{2})}{1+x^2}$$ has derivative equal to $$f'(x)=\frac{\exp(x/\sqrt{2})(1-2\sqrt{2}x+x^2)}{\sqrt{2}(1+x^2)^2}=\frac{\exp(x/\sqrt{2})((x-\sqrt{2})^2-1)}{\sqrt{2}(1+x^2)^2}$$ that vanishes at $x=-1+\sqrt{2}$ and $x=1+\sqrt{2}$. We are interested only in the first value since $-1+\sqrt{2}<1$ the other is bigger than one. On interval $[0,1]$ we get then that $f(x)$ is increasing for $x\in [0,\sqrt{2}-1)$ and decreasing for $x\in(\sqrt{2}-1,1]$ with the maximum at $x=\sqrt{2}-1$. By Mean Valued Theorem for integrals we have $$\int^1_0f(x)\,dx=f(\xi)(1-0)=f(\xi)$$ for some $\xi\in(0,1)$. By the above observations then we have $$f(\xi)=\int^1_0f(x)\,dx\leqslant f(\sqrt{2}-1)\int^1_0\,dx=f(\sqrt{2}-1)$$ and $$f(\xi)=\int^1_0f(x)\,dx\geqslant f(0)\int^1_0\,dx=f(0)=1$$

• I understand!! Could we get a better approximation if we use the second mean value theorem for integrals? This theorem is the following, isn't it? On $[a, b]$ let $f$ be monotone and $f'$ and $g$ be continuous. Then there exists a $\xi \in (a, b)$ with \begin{equation*}\int_a^bf(x)g(x)\, dx=f(a)\int_a^{\xi}g(x)\, dx+f(b)\int_{\xi}^bg(x)\, dx\end{equation*} Do we choose as $f(x):=\frac{1}{1+x^2}$ and $g(x):=e^{\frac{x}{\sqrt{2}}}$ ? – Mary Star Apr 7 '18 at 8:56
• @MaryStar I think you only need $f$ monotone and $g$ integrable. Anyways you could then find estimates for each of those two integrals. Maybe you get better bounds. – Arian Apr 7 '18 at 9:11
• Ah ok! So, do we chose as $f$ the function $\frac{1}{1+x^2}$ and as $g$ the function $e^{\frac{x}{\sqrt{2}}}$ or otherwise around? – Mary Star Apr 7 '18 at 9:19
• you can try both combinations, they are both monotonic and integrable on $[0,1]$. see which is better. – Arian Apr 7 '18 at 9:36
• Thank you!! :-) – Mary Star Apr 11 '18 at 21:59

The most obvious thing to do is to tkae $f(x)=\frac{e^{x/\sqrt2}}{1+x^2}$. It is then a matter of finding the sup and inf of $f$ on the interval $(0,1)$.

$$f'(x)=\frac{\Big(\frac{1}{\sqrt 2}(1+x^2)-(2x)\Big)e^\frac{x}{\sqrt 2}}{(1+x^2)^2}$$

The derivative is $0$ when $x=\sqrt 2 \pm 1$