# Is this continuous analogue to the AM–GM inequality true?

First let us remind ourselves of the statement of the AM–GM inequality:

Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $N\geqslant 1$ non-negative real numbers, we have $$\frac1N\sum_k x_k \geqslant \left(\prod_k x_k\right)^{\frac1N}$$

It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete sequences. This generalisation is the integral operator ∫.

Likewise, we can generalise the product operator to operate on continuous functions. We begin with the following property of the discrete product: $$\log\prod_k A_k=\sum_k \log A_k$$ (assuming that $A_k>0$). Using this we can define the continuous product (also known as the geometric integral according to Wikipedia) as follows: $$\prod_a^b f(x)\ dx := \exp\int_a^b\log f(x)\ dx.$$

assuming that the integral converges. From these definitions we can seek to generalise the AM–GM inequality to continuous functions:

Proposition: (Continuous AM–GM Inequality) For any suitably well-behaved non-negative function $f$ defined on $[a,b]$ with $a<b$, we have: $$\frac1{b-a}\int_a^bf(x)\ dx \geqslant \left(\prod_a^b f(x)\ dx\right)^\frac1{b-a}$$ or in traditional notation: $$\frac1{b-a}\int_a^bf(x)\ dx \geqslant \exp\left(\frac1{b-a} \int_a^b \log f(x)\ dx\right)$$

As a simple illustration, consider the function $f(x)=1+\sin x$ and fix the lower bound $a=0$. On the following graph the $x$-axis represents the upper bound of integration $b$, and on the $y$-axis are represented the the arithmetic mean (blue) and the geometric mean (red) of $f$ on the interval $[a,b]$ where $a=0$ and $b=x$.

Clearly the claimed inequality seems to hold, and is in fact quite tight when $b\approx0$.

This inequality is of interest not least because it is readily abe to produce a number of non-trivial numerical inequalities pertaining to known constants. For instance, again with the example $f(x)=1+\sin x$, set $b=\pi/2$. Then assuming the proposition holds, we have

$$\frac{2}{\pi}\int_0^{\pi/2} 1 + \sin x\ dx \geqslant \exp \left( \frac{2}{\pi} \int_0^{\pi/2} \log(1+\sin x)\ dx \right)$$ Evaluating these integrals and rearranging, we obtain: $$G \leqslant \frac\pi4\log\left(2+\frac2\pi\right) \approx 0.9313$$ where $G\approx 0.9159$ is Catalan's constant.

Anyway I would like to ask, firstly:

Is the claimed inequality true, and if so what is the proof and on what class of functions is it applicable?

Secondly, and this is more of a fun little challenge:

Assuming its veracity, can this inequality be used to prove remarkable numerical inequalities, e.g. $\pi < 22/7$ or $e^\pi - \pi < 20$?

• Have you heard about Jensen's inequality? Apr 23 '17 at 12:29
• Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it... Apr 23 '17 at 13:31
• Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer. Apr 23 '17 at 13:49
• Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $\Delta x$, similarly to how multiplying by $\Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration. Apr 23 '17 at 20:28
• Similarly, you can also consider \$\operatorname{HM}(f,a,b)=\frac{b-a}{\int_a^b\frac{dx}{f(x)}} as generalized harmonic mean. Apr 24 '17 at 17:26

I would write this inequality as $$\frac1{b-a}\int_a^b\exp(g(x))\,dx\ge\exp\left(\frac1{b-a}\int_a^b g(x)\,dx\right).$$ In disguise it is the case of Jensen's inequality for the convex function $$\phi(t)=\exp(t)$$.