Inequality between AM-GM

Hint: Make the substitution: $t = \dfrac{x}{y} > 1$, then the inequality you need to prove:

$$t^{1/2} \le \dfrac{t - 1}{\ln t} \le \frac{1+t}{2}$$

You can take their difference as a function of $t$, compute their derivative and show they are positive.

For $ 0<a<b$ note $ A(a,b)=\frac{a+b}{2}, G(a,b)=\sqrt{ab}, L(a,b)=\frac{b-a}{\ln b-\ln a}$

Proof 1.:

With $\frac{b}{a} = x > 1$ we have: $\sqrt{x}< \frac{x - 1}{\ln x} < \frac{1}{2}( x + 1)$

Let functions $f, g : [ 1, \infty) \rightarrow \textbf{R}$: $f( x ) = \ln x -\sqrt{x}+ \frac{1}{\sqrt{x}}, g( x ) = \ln x - \frac{2(x-1)}{x+1}$

with $f'( x ) = -\frac{(\sqrt{x}-1)^2}{2x\sqrt{x}}, g'( x ) = \frac{(x -1)^2}{x(x+1)}$ and $f(x) < 0 < g( x ),(\forall )x > 1 $.

Proof 2.:

Let $ H_{a,b}( x ): [ 0, 1 ]\rightarrow \textbf{R}$, $H_{a,b}( x ) = \frac{1}{2}( a^x.b^{1-x} + a^{1-x}.b^x )$ with $ L( a,b ) = \int^{1}_{0}H_{a,b}( x )dx, $ $H_{a,b}( 0 ) = H_{a,b}( 1 ) = A( a, b)$ and $H_{a,b}( \frac{1}{2} )= G( a, b ).$

$H^{'}_{a,b}( x ) = \frac{1}{2}( a^x.b^{1-x} - a^{1-x}.b^x )\ln \frac{b}{a}$ implies $G( a, b ) \leq H_{a,b}( x ) \leq A( a, b),(\forall) x\in [ 0, 1 ].$

In what follows we use other depictions of logarithmic mean using integral:

$\frac{1}{L( a, b )} = \frac{1}{b - a}\int^{b}_{a}\frac{dx}{x}=\frac{1}{b - a}\int^{\frac{b}{a}}_{1}\frac{dx}{x}$

$\frac{1}{L( a, b )} = \int^{1}_{0}\frac{dx}{xa + ( 1- x )b}$

$\frac{1}{L( a, b )} = \lim_{x\rightarrow 0}\int^{x}_{0}\frac{dt}{(t + a)( t + b )}$

Proof 3.:

For $(\forall)x > 0$ we have: $\frac{2}{(x+1)^2}\leq\frac{1}{2x}\leq\frac{x+1}{4x\sqrt{x}}$

Integrating this inequality from $1$ to $\frac{b}{a}$ now obtain :

$\frac{b-a}{b+a}\leq\frac{1}{2}ln\frac{b}{a} \leq\frac{b-a}{2\sqrt{ab}}...$

Proof 4.:

For $(\forall)x \geq 0$ we have:

$x^{2} + 2x\sqrt{ab} + ab \leq x^{2} + x(a + b) + ab \leq x^{2} + x(a + b) + (\frac{a + b}{2})^2$

Reversing these inequalities and integrating them using last representation is obtained inequality.

Proof 5.:

Noting $a = e^{x}, b = e^{y}$ demonstrated inequality may be written as

$e^{\frac{x+y}{2}} \leq \frac{e^y - e^x}{y - x} \leq \frac{e^x + e^y}{2}.$

Dividing this inequality with $e^{\frac{x+y}{2}}$the first part to write:

$1\leq \frac{e^{\frac{y - x}{2}} - e^{\frac{x - y}{2}}}{y - x}\Longleftrightarrow \frac{y - x}{2} \leq \sinh\frac{y -x}{2}$ true that $t \leq \sinh t,( \forall) t\geq 0.$

Similarly, the second part is reduced to $t \geq \tanh t,( \forall) t\geq 0.$

Proof 6.:

Inequality in the previous proof can be obtained by applying Hadamard-Hermite's inequality function $f: \textbf{R}\rightarrow \textbf{R}, f( x ) = e^{x}$

(Hadamard-Hermite's inequality: $f ( \frac{a + b}{2} )\leq\frac{1}{b - a}.\int^{b}_{a}f( x ) dx \leq \frac{f ( a ) + f ( b )}{2}.$)

Inequality of utterance can be demonstrated using CBS with integrals of convenient functions chosen.

Proof 7.:

If $f:[a, b]\rightarrow(0, \infty)$ then

$\int^{b}_{a}(\sqrt{f(x)})^2dx.\int^{b}_{a}\frac{1}{(\sqrt{f(x)})^2}dx\geq(\int^{b}_{a}(\sqrt{f(x)})^2.\frac{1}{(\sqrt{f(x)})^2}dx)^2$ and

$\int^{b}_{a}f(x)dx.\int^{b}_{a}\frac{1}{f(x)}dx\geq (b-a)^2.$

Now for $f(x)=\frac{1}{x}$ we have $(\ln b-\ln a).\frac{b^2-a^2}{2}>(b-a)^2$ ie $L ( a, b )\leq A(a, b).$

For $f(x)=e^x$ is obtained $(e^b-e^a)(e^{-a}-e^{-b})>(b-a)^2$ and noting $a=\ln x, b=\ln y$ cu $0<x<y$ result $G(x,y)< L(x,y).$

Prof 8.:

If $f:[a, b]\rightarrow(0, \infty)$ then $\int^{b}_{a}(f(x))^2dx.\int^{b}_{a}1dx\geq(\int^{b}_{a}f(x).1dx)^2$ or $(b-a)\int^{b}_{a}(f(x))^2dx\geq(\int^{b}_{a}f(x)dx)^2.$ Now, applying this inequality for $f(x)=\frac{1}{x}$ is obtained

$(\ln b- \ln a)^2<(b-a)(\frac{1}{a}-\frac{1}{b})$ from which it follows that $G(a,b)< L(a,b).$

For $f(x)=\frac{1}{\sqrt{x}}$ rezult $4(\sqrt{b}-\sqrt{a})^2<(b-a)(\ln b-\ln a).$ Noting $\sqrt{b}=y$ and $\sqrt{a}=x$ we have $4(y-x)<2(y^2-x^2)(\ln y-\ln x)$and finally $L(x,y)<A(x,y).$

(Paper presented at the XVIII Conference of the Mathematical Society of Romania, Iasi, 24 -26 October 2014)