Prove inequality with MVT

Question

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*}

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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?

Answer 1

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 $$f(\xi)=\int^1_0f(x)\,dx\geqslant f(0)\int^1_0\,dx=f(0)=1$$

Answer 2

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 $x=\sqrt 2 \pm 1$