An Inequality with Just Two Variables II

Problem

An  Inequality with Just Two Variables II

Solution 1

$\displaystyle\begin{align} LHS &= 3+\frac{4\sqrt{ab}}{a+b}+\frac{a+b}{\sqrt{ab}}+\frac{4ab}{(a+b)^2}+\frac{(a+b)^2}{4ab}\\ &\le 3+2+\frac{a+b}{\sqrt{ab}}+1+\frac{(a+b)^2}{4ab}\\ &= 6+\frac{a+b}{\sqrt{ab}}+\frac{(a+b)^2}{4ab}. \end{align}$

Suffice it to show that

$\displaystyle 6+\frac{a+b}{\sqrt{ab}}+\frac{(a+b)^2}{4ab}\le 5 +2 \left(\frac{a}{b}+\frac{b}{a}\right)$

which is equivalent to

(1)

$\displaystyle 1+\frac{a+b}{\sqrt{ab}}+\frac{(a+b)^2}{4ab}\le \frac{2(a^2+b^2)}{ab}.$

Now, by the GM-HM inequality, $\displaystyle\sqrt{ab}\ge\frac{2ab}{a+b},\;$ hence $\displaystyle\frac{a+b}{\sqrt{ab}}\le\frac{(a+b)^2}{2ab}.$

It follows that

(2)

$\displaystyle \frac{a+b}{\sqrt{ab}}+1+\frac{(a+b)^2}{4ab}\le 1+\frac{3(a+b)^2}{4ab}.$

(1) and (2) show that the problem will be solved if we manage to prove

$\displaystyle\frac{3(a+b)^2+4ab}{4ab}\le \frac{2(a^2+b^2)}{ab}.$

This is equivalent to $3a^2+3b^2+10ab\le 8a^2+8b^2,\,$ or $5a^2+5b^2-10ab\ge 0,\;$ which is simply $5(a-b)\ge 0.$

Solution 2

We employ the obvious $\displaystyle\frac{2\sqrt{ab}}{a+b}\le 1\,$ and $\displaystyle\frac{4ab}{(a+b)^2}\le 1\,$ to obtain

$\displaystyle LHS\le 6+\frac{3(a+b)^2}{4ab}.$

Suffice it to prove that $\displaystyle 6+\frac{3(a+b)^2}{4ab}\le 5+2\left(\frac{a}{b}+\frac{b}{a}\right).$

Now note that

$\displaystyle 1+\frac{3(a+b)^2}{4ab}\le\frac{5}{2}+\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)\le 2\left(\frac{a}{b}+\frac{b}{a}\right),$

Because $\displaystyle\frac{5}{2}\le\frac{5}{4}\left(\frac{a}{b}+\frac{b}{a}\right).$

Solution 3

Using the AM-GM inequality, $\displaystyle\frac{a+b}{2}\ge\sqrt{ab}\,$ and $\displaystyle\frac{2}{a+b}\le\frac{1}{\sqrt{ab}},$

$\displaystyle\begin{align} LHS &\le \left(\sqrt{ab}+\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{a+b}{2ab}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}}\right)\\ &= \left(2\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{a+b}{2ab}+\frac{2}{\sqrt{ab}}\right)\\ &=\frac{a+b}{\sqrt{ab}}+4+\frac{(a+b)^2}{4ab}+\frac{a+b}{\sqrt{ab}}\\ &= 4+\frac{(a+b)^2}{4ab}+\frac{2(a+b)}{\sqrt{ab}}. \end{align}$

Thus, suffice it to prove that

$\displaystyle 4+\frac{(a+b)^2}{4ab}+\frac{2(a+b)}{\sqrt{ab}}\le 5 +2 \left(\frac{a}{b}+\frac{b}{a}\right),$

which is equivalent to

$\displaystyle \frac{(a+b)^2}{4ab}+\frac{2(a+b)}{\sqrt{ab}}\le 1 +2\frac{a^2+b^2}{ab}=\frac{2a^2+2b^2+ab}{ab}.$

Now, from $\displaystyle\frac{1}{\sqrt{ab}}\le\frac{a+b}{2ab}\,$ we get $\displaystyle\frac{2(a+b)}{\sqrt{ab}}\le\frac{(a+b)^2}{ab},\,$ implying

$\displaystyle \frac{(a+b)^2}{4ab}+\frac{2(a+b)}{\sqrt{ab}}\le \frac{(a+b)^2}{4ab}+\frac{(a+b)^2}{ab}=\frac{5(a+b)^2}{4ab}.$

Thus, suffice it to prove $\displaystyle\frac{5}{4}\frac{(a+b)^2}{ab}\le\frac{2a^2+2b^2+ab}{ab},\,$ i.e.,

$\displaystyle 5a^2+5b^2+10ab\le 8a^2+8b^2+4ab$

which reduces to $3(a-b)^2\ge 0.$

Solution 4

Let

$\displaystyle\begin{align} A &= \left(\frac{2ab}{a+b}+\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{a+b}{2ab} + \frac{1}{\sqrt{ab}}+\frac{2}{a+b}\right)\\ &=\left(\frac{2ab}{a+b}+\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{2}{a+b} + \frac{1}{\sqrt{ab}}+\frac{a+b}{2ab}\right)\\ &=\frac{1}{ab}\left(\frac{2ab}{a+b}+\sqrt{ab}+\frac{a+b}{2}\right)^2\\ &\le\frac{1}{ab}(\sqrt{ab}+a+b)^2\\ &=\frac{1}{ab}(ab+2\sqrt{ab}(a+b)+(a+b)^2)\\ &\le\frac{1}{ab}(ab+2(a+b)^2)\\ &=\frac{1}{ab}(5ab+2(a^2+b^2))\\ &=5+2\left(\frac{a}{b}+\frac{b}{a}\right). \end{align}$

Solution 5

Define $\displaystyle x=\frac{2ab}{a+b},\,$ $\displaystyle y=\sqrt{ab},\,$ $\displaystyle z=\frac{a+b}{2}.\,$ We have $x\le y\le z\,$ and $y^2=xz.

$\displaystyle\begin{align} LHS &= \left(\sum_{cycl}x\right)\left(\sum_{cycl}\frac{1}{x}\right)\\ &=\frac{\left(\displaystyle\sum_{cycl}x\right)(xy+yz+zx)}{xyz}\\ &=\frac{\left(\displaystyle\sum_{cycl}x\right)(xy+yz+y^2)}{xyz}\\ &=\frac{(x+y+z)^2}{xz}\\ &=\frac{(x+y+z)^2}{y^2}. \end{align}$

$\displaystyle\begin{align} RHS &= \frac{2a^2+2b^2+5ab}{ab}\\ &= \frac{2(a+b)^2+ab}{ab}\\ &=1+\frac{8z^2}{y^2}. \end{align}$

The required inequality is equivalent to $\displaystyle\frac{(x+y+z)^2}{y^2}\le 1+\frac{8z^2}{y^2}.\,$ which is $(x+y+z)^2-y^2\le 8z^2,\,$ or, $(x+2y+z)(x+z)\le 8z^2.$

Since $x\le y \le z,\,$ $x+2y+z\le 4z\,$ and $x+z\le 2z,\,$ which shows that, indeed, $(x+2y+z)(x+z)\le 8z^2.$

Solution 6

We know that, for positive $a,b,c,$

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

We'll use Schweitzer's inequality:

$\displaystyle\left(\sum_{k=1}^{n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\le\frac{(m+M)^2n^2}{4mM},$

where $x_1,\ldots,x_n\in [m,M],\;$ $m\gt 0.\;$

with $n=3,\;$ $\displaystyle m=x_1=\frac{2ab}{a+b},\;$ $\displaystyle x_2=\sqrt{ab},\;$ and $x_3=\displaystyle\frac{a+b}{2}=M,\;$ we directly get

$\displaystyle\begin{align} A&=\left(\frac{2ab}{a+b}+\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{a+b}{2ab} + \frac{1}{\sqrt{ab}}+\frac{2}{a+b}\right)\\ &\le 9\frac{\displaystyle\left(\frac{2ab}{a+b}+\frac{a+b}{2}\right)^2}{4ab}\\ &= \frac{9}{4ab}\left[\left(\frac{2ab}{a+b}\right)^2+2ab+\left(\frac{a+b}{2}\right)^2\right]\\ &\le \frac{9}{4ab}\left[(\sqrt{ab})^2+2ab+\left(\frac{a+b}{2}\right)^2\right]\\ &= \frac{9}{4ab}\left[3ab+\left(\frac{a+b}{2}\right)^2\right]\\ &= \frac{9}{16ab}\left[14ab+(a^2+b^2)\right]\\ &=\frac{63}{8}+\frac{9}{16}\left(\frac{a}{b}+\frac{b}{a}\right). \end{align}$

Now, $\displaystyle 1\le\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}\right).\,$ It then follows that $\displaystyle\frac{23}{8}\le \frac{23}{16}\left(\frac{a}{b}+\frac{b}{a}\right)$ and, subsequently,

$\displaystyle \frac{63}{8}+\frac{9}{16}\left(\frac{a}{b}+\frac{b}{a}\right)\le\frac{40}{8}+\frac{32}{16}\left(\frac{a}{b}+\frac{b}{a}\right)= 5+2\left(\frac{a}{b}+\frac{b}{a}\right).$

Acknowledgment

The problem above has been kindly posted to the CutTheKnotMath facebook page by Dan Sitaru, along with several solutions. Solutions 1,3,5 are by Soumava Chakraborty; Solution 2 by Kevin Soto Palacios; Solution 4 is by Abdallah El Farissi. Solution 6, making use of Schweitzer's inequality, highlighted by Dan Sitaru elsewhere, yields a somewhat better estimate.

 

Inequalities in Two Variables

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