Can we prove AM-GM Inequality using these integrals? I came across these two results recently:
$$ \int_a^b \sqrt{\left(1-\dfrac{a}{x}\right)\left(\dfrac{b}{x}-1\right)} \: dx = \pi\left(\dfrac{a+b}{2} - \sqrt{ab}\right)$$
$$ \int_a^c \sqrt[3]{\left| \left(1-\dfrac{a}{x}\right)\left(1-\dfrac{b}{x}\right)\left(1-\dfrac{c}{x}\right)\right|} \: dx = \dfrac{2\pi}{\sqrt{3}}\left(\dfrac{a+b+c}{3} - \sqrt[3]{abc}\right)$$
for $0<a\leq b\leq c$.
I haven't tried to solve the first one yet, but I have an idea of how to approach it, namely using the substitution $x=a\cos^2\theta+b\sin^2\theta$. I have no idea how to approach the second one, however.
I think that the most interesting thing about the results above is that it seems like there is a proof for the AM-GM inequality hidden within. Clearly both integrands are positive and so the AM-GM falls out for the 2 and 3 variable case. All that is required is to prove the results.
My question is twofold:

*

*How would the second integral be computed? Is there an approach using elementary techniques?

*Can this be generalised to prove the AM-GM inequality for $n$-variables?

 A: $\DeclareMathOperator{\Res}{Res}$
$\DeclareMathOperator{\sgn}{sgn}$
$\newcommand{\d}{\mathrm{d}}$
$\newcommand{\e}{\mathrm{e}}$
$\newcommand{\E}{\mathrm{E}}$
$\newcommand{\P}{\mathrm{P}}$
$\newcommand{\i}{\mathrm{i}}$
$\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}$
Let $0 < a_0 \leq a_1 \ldots \leq a_{n-1}$ be a monotone sequence of $n$
positive real numbers. Then $$\boxed{
  \sum_{k=0}^{n-2}\frac{1}{\pi}\sin \tfrac{\pi (k+1)}{n}\int_{a_k}^{a_{k+1}}
\left(\prod_{i=0}^{n-1}\sqrt[n]{\abs{x-a_i}}\right)\frac{\d x}{x}
  =\frac{1}{n}\sum_{i=0}^{n-1}a_i - \sqrt[n]{\prod_{i=0}^{n-1}a_i}
}\text{.}$$
The left side is manifestly nonnegative, so OP's second question is answered in the affirmative, although the form of the left side for $n=2,3$ is deceptively simple and does not reflect the general case.
The underlying method of this answer does not differ from that of Svyatoslav's, save for doing the bookkeeping needed to provide for the $n$-variable case. The method can be adapted to show that, for a positive random variable $X$ that takes on a finite set of values,
$$
  \int_{\mathrm{ess}\,\inf X}^{\mathrm{ess}\,\sup X}
\frac{\sin (\pi \P(X >x))}{\pi x}\e^{\E \ln \abs{X-x}} \d x
  =\E\,X - \e^{\E\ln X}
\text{;}$$
it's tempting to think that this last equality holds for arbitrary positive random variables with finite arithmetic and geometric mean, but I have no proof of that.

Write $I$ for the closed interval $[a_0,a_{n-1}]$. Take
the cut of $\sqrt[n]{z}$ to be the negative real axis. Write $\mathbb{C}^*$ for the Riemann sphere. Consider the meromorphic
differential forms $\alpha_{U_0}$, $\alpha_{U_1}$, $\alpha_{U_2}$,
on $U_0 =
\mathbb{C}\backslash (-\infty,a_{n-1}]$, $U_1 = \mathbb{C}\backslash
[a_0,\infty)$, and $U_2=\mathbb{C}^*\backslash [0,a_{n-1}]$ given by
$$\begin{aligned}
  \alpha_{U_0} &=\left(\prod_{i=0}^{n-1}\sqrt[n]{z-a_i}\right)\frac{\d z}{z} \\
  \alpha_{U_1} &=-\left(\prod_{i=0}^{n-1}\sqrt[n]{a_i-z}\right)\frac{\d z}{z} \\
  \alpha_{U_2} &=-\left(\prod_{i=0}^{n-1}\sqrt[n]{1-a_iz^{-1}}\right)\frac{\d (z^{-1})}{(z^{-1})^2}\text{.}
\end{aligned}$$
These three forms agree pairwise on the intersections of their respective
domains. That's because—for the chosen cut convention—

*

*if $a > 0$ and either $\Im z \neq 0$ or $\Re z > a$, then
$$\sqrt[n]{z-a} = \frac{\sqrt[n]{1-az^{-1}}}{\sqrt[n]{z^{-1}}}\text{;}$$


*if $a > 0$ and $\Im z \neq 0$, then
$$\sqrt[n]{a-z} = \e^{-\i\pi \sgn \Im z /n}\sqrt[n]{z-a}\text{.}$$
Consequently, there is a unique meromorphic differential form $\alpha$ on
$\mathbb{C}^*\backslash I = \bigcup_{i=0}^2U_i$ such that $\left.
\alpha\right\rvert_{U_i} = \alpha_{U_i}$ for $i=0,1,2$. This $\alpha$ has a
simple pole at $0$ and a double pole at $\infty$.
So let $C$ be a cycle separating $I$ from $\{0,\infty\}$ and oriented in the
negative sense. What is
$$\frac{1}{2\pi\i}\oint_C\alpha\text{?}$$

*

*If $C$ is taken to be a rectangle with sides parallel to and infinitesimally
close to $I$, then
$$\frac{1}{2\pi\i}\oint_C \alpha
    = \frac{1}{2\pi\i}\left(\int_{I+\i 0^+}\alpha -
    \int_{I-\i0^+}\alpha\right)\text{,}$$
the contribution from the remaining sides vanishing.


*If $C$ is taken to encircle $\{0,\infty\}$ in a positive sense, then
$$\frac{1}{2\pi\i}\oint_C \alpha
    = \Res_0 \alpha + \Res_{\infty}\alpha\text{.}$$
For the former choice, note that, for real $x$,
$$\sqrt[n]{x\pm \i 0^+} = \sqrt[n]{\abs{x}}\e^{\pm\i\pi [x < 0]/n}$$
where $[(-)]$ is Iverson bracket notation.
Therefore
$$\int_{I\pm\i0^+}\alpha = \int_I\e^{\pm\i\pi
N(x)/n}\left(\prod_{i=0}^{n-1}\sqrt[n]{\abs{x-a_i}}\right)\frac{\d x}{x}$$
where $N(x)$ is the number of the $a_i$ greater than $x$. Then
$$\frac{1}{2\pi\i}\oint_C\alpha = \frac{1}{\pi}\int_I\sin \tfrac{\pi N(x)}{n}
\left(\prod_{i=0}^{n-1}\sqrt[n]{\abs{x-a_i}}\right)\frac{\d x}{x}$$
which, because $N(x)$ is constant away from the $a_i$, simplifies to
$$\frac{1}{2\pi\i}\oint_C\alpha =
  \sum_{k=0}^{n-2}\frac{1}{\pi}\sin \tfrac{\pi (k+1)}{n}\int_{a_k}^{a_{k+1}}
\left(\prod_{i=0}^{n-1}\sqrt[n]{\abs{x-a_i}}\right)\frac{\d x}{x}\text{.}$$
As for the latter choice: from the Laurent expansions of $\alpha$ at $0$, $\infty$
$$\alpha_z = \left(-\sqrt[n]{\prod_{i=0}^{n-1}a_i} + \mathcal{O}(z)\right)\frac{\d z}{z}$$
$$\alpha_z = \left(-\frac{1}{z^{-1}} + \frac{1}{n}\sum_{i=0}^{n-1}a_i + \mathcal{O}(z^{-1}) \right)\frac{\d (z^{-1})}{z^{-1}}$$
the required residues are found to be
$$\Res_0 \alpha = - \sqrt[n]{\prod_{i=0}^{n-1}a_i}$$
$$\Res_{\infty} \alpha = \frac{1}{n}\sum_{i=0}^{n-1}a_i$$
whence
$$\frac{1}{2\pi\i}\oint_C\alpha =
\frac{1}{n}\sum_{i=0}^{n-1}a_i - \sqrt[n]{\prod_{i=0}^{n-1}a_i}\text{.}$$
These two choices of $C$ must result in the same value for $\tfrac{1}{2\pi\i}\int_C\alpha$, whence
$$\boxed{
  \sum_{k=0}^{n-2}\frac{1}{\pi}\sin \tfrac{\pi (k+1)}{n}\int_{a_k}^{a_{k+1}}
\left(\prod_{i=0}^{n-1}\sqrt[n]{\abs{x-a_i}}\right)\frac{\d x}{x}
  =\frac{1}{n}\sum_{i=0}^{n-1}a_i - \sqrt[n]{\prod_{i=0}^{n-1}a_i}
}\text{.}$$
A: $$I(a,b,c)=\int_a^c\Big|(x-a)(x-b)(x-c)\Big|^{1/3}\frac{dx}{x}=I_{ab}+I_{bc}$$
where $I_{ab}=\int_a^b$, etc.
We go to the complex plane and consider $\oint$ with the turning points $z=a$ and $z=c$ (making the cut from $a$ to $c$). Let's denote this integral as $j$. It is easy to check that
$$j=\oint=I_{ab}+I_{bc}e^{-\pi i/3}+I_{cb}e^{-\pi i}+I_{ba}e^{-4\pi i/3}$$
where $I_{ba}=-I_{ab}$ (we go in the opposite direction - from $b$ to $a$ in the lower bank of the cut). We also added small half- and full circles around $z=a,b,c$ - to make the contour closed (the integrals along these circles have zero contribution).
Now, we add the integration path from $z=c$ to $R$ along axis $X$, and a big circle of the radius $R\to \infty$:
we integrate along the upper bank of the cut, then from $z=c$ to $R$, along a big circle, then from $z=R$ back to $z=c$ along axis $X$ in the opposite direction, and finally  along the lower bank of the cut to the starting point. Let's denote this integral as $J$.
As we have the only pole inside this big closed contour, and bearing in mind that integrals along axis $X$ cancel each other, we can write:
$$J=j+\oint_R=2\pi i \operatorname{Res}\Big((z-a)(z-b)(z-c)\Big)^{1/3}\frac{1}{z}$$
where $\oint_R=\oint_R\Big((z-a)(z-b)(z-c)\Big)^{1/3}\frac{dz}{z}$ - denotes the integration along a big circle (counter-clockwise).
All evaluations are straightforward (the only thing is that we have to appoint correctly the phases of the integrand when going around $z=a,b,c$ on the both banks of the cut).
$$\operatorname{Res}_{z=0}\Big((z-a)(z-b)(z-c)\Big)^{1/3}\frac{1}{z}=e^{\pi i/3}(abc)^{1/3}$$
$$\oint_R\to 2\pi ie^{-2\pi i/3}\Big(-\frac{1}{3}\Big)(a+b+c)\,\, \text{at} \,R\to\infty$$
Taking all together
$$(I_{ab}+I_{bc})(1+e^{-\pi i/3})=2\pi ie^{\pi i/3}\Big((abc)^{1/3}-\frac{a+b+c}{3}\Big)$$
$$I(a,b,c)=\frac{\pi}{\cos(\pi /6)}\Big(\frac{a+b+c}{3}-(abc)^{1/3}\Big)$$
The pattern for $0<a<b<c<d<e...$ is also clear.
