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.

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