## Applications of Midpoint Convexity: AM-GM Inequalities

We proved in the last post that a test for the convexity of continuous function $f: \mathbb{R}\rightarrow\mathbb{R}$ is that $f(x+h)+f(x-h)-2f(x)\geq 0$, for all $x,h\in\mathbb{R}$. I claim that we can use this test to show that $f(x):=e^{x}$ is strictly convex. Observe that $\frac{a+b}{2}>\sqrt{ab}$, for $a,b > 0$ and $a\neq b$, by considering the quadratic $(\sqrt{a}-\sqrt{b})^{2}>0$. Hence,

$\displaystyle e^{x+h}+e^{x-h}-2e^{x}=e^{x+h}+e^{x-h}-2e^{x+h}e^{x-h}>0$,

for all $x\in\mathbb{R}$ and $h>0$. We use the strict convexity of $e^{x}$ to prove a weighted form of the arithmetic mean-geometric mean (AM-GM) inequality.

Proposition 1. If $x_{1},\cdots,x_{n}\in(0,\infty)$ and $\lambda_{1},\cdots,\lambda_{n}\in(0,1)$ such that $\sum_{k=1}^{n}\lambda_{k}=1$, then

$\displaystyle\sum_{k=1}^{n}\lambda_{k}x_{k}>x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}$,

unless $x_{1}=\cdots=x_{n}$.

Proof. Fix $x_{1},\cdots,x_{n}\in(0,\infty)$ and $\lambda_{1},\cdots,\lambda_{n}\in(0,1)$ such that $\sum_{k=1}^{n}\lambda_{k}=1$.

$\begin{array}{lcl}\displaystyle x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}=e^{\lambda_{1}\log x_{1}}\cdots e^{\lambda_{n}\log x_{n}}&=&\displaystyle e^{\sum_{k=1}^{n}\lambda_{k}\log x_{k}}\\&\leq&\displaystyle\sum_{k=1}^{n}\lambda_{k}e^{\log x_{k}}\\&=&\displaystyle\sum_{k=1}^{n}\lambda_{k}x_{k}\end{array}$

The necessity and sufficiency of the strictness condition is immediate from the strict convexity of $e^{x}$. $\Box$

Suppose $x_{1},\cdots,x_{n}$ and $\lambda_{1},\cdots,\lambda_{n}$ satisfy the same hypotheses as above. Then replacing $x_{k}$ by $x_{k}^{-1}$, for $1\leq k \leq n$, we obtain that

$\displaystyle\sum_{k=1}^{n}\dfrac{\lambda_{k}}{x_{k}}>\dfrac{1}{x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}}\Longleftrightarrow x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}>\left(\sum_{k=1}^{n}\dfrac{\lambda_{k}}{x_{k}}\right)^{-1}$

Lemma 2. Let $f: I \rightarrow \mathbb{R}$ be a convex function and $x_{1},\cdots,x_{n}\in I$ ($n\geq 2$). Then

$\displaystyle(n-1)\left[\dfrac{f(x_{1})+\cdots+f(x_{n-1})}{n-1}-f\left(\dfrac{x_{1}+\cdots+x_{n-1}}{n-1}\right)\right]\leq n\left[\dfrac{f(x_{1})+\cdots+f(x_{n})}{n}-f\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)\right]$

Proof. Observe that

$\begin{array}{lcl}\displaystyle f\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)&=&\displaystyle f\left(\dfrac{n-1}{n}\dfrac{x_{1}+\cdots+x_{n-1}}{n-1}+\dfrac{x_{n}}{n}\right)\\[.7 em]&\leq &\displaystyle\dfrac{n-1}{n}f\left(\dfrac{x_{1}+\cdots+x_{n-1}}{n-1}\right)+\dfrac{1}{n}f(x_{n})\end{array}$

Hence,

$\begin{array}{lcl}\displaystyle n\left[\dfrac{f(x_{1})+\cdots+f(x_{n})}{n}-f\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)\right]&\geq&\displaystyle f(x_{1})+\cdots+f(x_{n})-(n-1)f\left(\dfrac{x_{1}+\cdots+x_{n-1}}{n-1}\right)-f(x_{n})\\[.7 em]&=&\displaystyle (n-1)\left[\dfrac{f(x_{1})+\cdots+f(x_{n-1})}{n-1}-f\left(\dfrac{x_{1}+\cdots+x_{n-1}}{n-1}\right)\right]\end{array}$

$\Box$

We use the preceding lemma to prove more inequalities involving arithmetic and geometric means. For $n\geq 2$ and $x_{1},\cdots,x_{n}>0$, define

$\displaystyle A_{k}:=\dfrac{x_{1}+\cdots+x_{k}}{k},G_{k}:=(x_{1}\cdots x_{k})^{\frac{1}{k}},\indent\forall k\in\mathbb{Z}^{\geq 1}$

If we apply the above lemma to $f=-\log$, then

$\begin{array}{lcl}\displaystyle n\left[\dfrac{f(x_{1})+\cdots+f(x_{n})}{n}-f\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)\right]&=&\displaystyle n\left[\log\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)-\dfrac{\log(x_{1})+\cdots+\log(x_{n})}{n}\right]\\[.7 em]&=&\displaystyle n\left[\log\left(A_{n}\right)-\log\left(G_{n}\right)\right]\\[.7 em]&=&\displaystyle\log\left(\dfrac{A_{n}}{G_{n}}\right)^{n}\end{array}$

Hence,

$\displaystyle\log\left(\dfrac{A_{n}}{G_{n}}\right)^{n}\geq\log\left(\dfrac{A_{n-1}}{G_{n-1}}\right)^{n-1}\geq\cdots\geq\log\left(\dfrac{A_{1}}{G_{1}}\right)=0$

The monotonicity of the exponential implies the inequalities

$\displaystyle\left(\dfrac{A_{n}}{G_{n}}\right)^{\frac{1}{n}}\left(\dfrac{A_{n-1}}{G_{n-1}}\right)^{n-1}\geq\cdots\geq\left(\dfrac{A_{1}}{G_{1}}\right)=1$

Now let $f=\exp$. Then

$\begin{array}{lcl}\displaystyle n(A_{n}-G_{n})=n\left[\dfrac{x_{1}+\cdots+x_{n}}{n}-\left(x_{1}\cdots x_{n}\right)^{\frac{1}{n}}\right]&=&\displaystyle n\left[\dfrac{e^{\log x_{1}}+\cdots+e^{\log x_{n}}}{n}-\left(e^{\log x_{1}}\cdots e^{\log x_{n}}\right)^{\frac{1}{n}}\right]\\[.9 em]&=&\displaystyle n\left[\dfrac{e^{\log x_{1}}+\cdots+e^{\log x_{n}}}{n}-\exp\left(\dfrac{\log x_{1}+\cdots+\log x_{n}}{n}\right)\right]\\[.9 em]&=&\displaystyle n\left[\dfrac{f(\log x_{1})+\cdots+f(\log x_{n})}{n}-f\left(\dfrac{\log x_{1}+\cdots+\log x_{n}}{n}\right)\right]\end{array}$

Hence,

$\displaystyle n(A_{n}-G_{n})\geq n(A_{n-1}-G_{n-1})\geq\cdots\geq 1 \left(A_{1}-G_{1}\right)=0$

The preceding two inequalities are apparently due to T. Popoviciu and R. Rado, respectively.