Inequality with Two Variables? Think Again

Preliminaries

In response to a recent page concerned with the problem of proving

$\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}.$

which holds for $a,b\in [0,1],\,$ an inequality in two variables, Professor Dorin Marghidanu email to me an issue of the Romanian Magazine Cardinal (Anul XXV , Nr. 2, 2014/2015), with an article (pp 1-5) coauthored by Leo Giugiuc, which, in particular, extends the above problem to any number of variables and, in addition, establishes an extra inequality on the "other side." The article treats the inequality for what it is: the relationship between various averages.

For non-negative $a_1,a_2,\ldots,a_n,\,$ $n\in\mathbb{N},\,$ $n\ge 1,\,$ the article uses the following notations:

$\displaystyle \begin{align} A_n[a]:=\frac{1}{n}\sum_{k=1}^na_n,\,\text{the arithmetic mean of},\,a_1,a_2,\ldots,a_n\\ G_n[a]:=\sqrt{\prod_{k=1}^na_n},\,\text{the geometric mean of},\,a_1,a_2,\ldots,a_n \end{align}$

Problem

Inequality with Two Variables? Think Again

Solution

Consider the function $f:\,(0,\infty)\rightarrow\mathbb{R},\,$ defined by $\displaystyle f(x)=\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}.\,$ Easily, $f'(x)=-\displaystyle \frac{1}{2}(1+x)^{-\frac{3}{2}}\lt 0,\,$ $\displaystyle f''(x)=\frac{3}{4}(1+x)^{-\frac{5}{2}}\gt 0,\,$ which tells us that $f(x)\,$ is convex on $(0,\infty).$

Applying Jensen's inequality, we have

$\displaystyle \begin{align} &\frac{1}{n}\sum_{k=1}^nf(a_k)\ge f\left(\frac{1}{n}\sum_{k=1}^{n}a_k\right)\,\Longleftrightarrow\\ &\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\ge\frac{1}{\displaystyle \sqrt{1+\frac{1}{n}\sum_{k=1}^na_k}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\ge\frac{n}{\sqrt{1+A_n[a]}}. \end{align}$

Equality holds when $a_1=a_2=\ldots=a_n.$ This proves the left inequality on $(0,\infty).\,$

To prove the right inequality, make the substitution $b_k=\ln a_k,\,$ $k=1,2\ldots,n.\,$ Then from $a_k\in (0,2),\,$ $b_k\in(-\infty,\ln 2)\,$ and $\displaystyle \sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}=\sum_{k=1}^n\frac{1}{\sqrt{1+e^{b_k}}}.$

Let's consider the function $g:\,(-\infty,\ln 2)\rightarrow\mathbb{R},\,$ defined by $\displaystyle g(x)=\frac{1}{\sqrt{1+e^x}}.\,$ We have $\displaystyle g'(x)=-\frac{1}{2}\cdot (1+e^x)^{-\frac{3}{2}}\cdot e^x,\,$ $g''(x)=\frac{1}{4}\cdot (1+e^x)^{-\frac{5}{2}}\cdot e^x(2-e^x)\lt 0,\,$ which shows that the function is concave for $x\lt\ln 2.\,$ Again invoking Jensen's inequality, we have

$\displaystyle \begin{align} &\frac{1}{n}\sum_{k=1}^ng(a_k)\le g\left(\frac{1}{n}\sum_{k=1}^{n}b_k\right)\,\Longleftrightarrow\\ &\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{1+e^{b_k}}}\le\frac{1}{\displaystyle \sqrt{1+e^{\frac{1}{n}\sum_{k=1}^nb_k}}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+e^{b_k}}}\le\frac{n}{\sqrt{1+e^{\frac{1}{n}\sum_{k=1}^nb_k}}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\le\frac{n}{\sqrt{1+e^{\frac{1}{n}\sum_{k=1}^n\ln a_k}}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\le\frac{n}{\sqrt{1+e^{\frac{1}{n}\ln\left(\prod_{k=1}^n a_k\right)}}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\le\frac{n}{\sqrt{1+e^{\displaystyle \ln G_n}}}\,\Longleftrightarrow\\ &\sum_{k=1}^n\frac{1}{\sqrt{1+a_k}}\le\frac{n}{\sqrt{1+G_n}}. \end{align}$

Equality holds when $a_1=a_2=\ldots=a_n.$ This proves the right inequality on $(0,2).\,$

Remark

Quite obviously, with $n=2,\,$ $a_1=a^2\,$ and $a_2=b^2,\,$ the right inequality resolves the earlier problem.

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