Comparing two close numbers How to compare these two numbers without using a calculator ?
$A=\left(\dfrac{11}{10}\right)^{\sqrt{5}}$ and  $\;B=\left(\dfrac{12}{11}\right)^{\sqrt{6}}$.
Thanks for your help !
Here is what I tried for example : $$\left(\frac{A}{B}\right)^{\sqrt6-\sqrt5}=\frac{11}{10^{\sqrt{30}−5}12^{6−\sqrt{30}}}.$$
ln is concave, so $$10^{\sqrt{30}−5}12^{6−\sqrt{30}}\leq10(\sqrt{30}−5))+12(6−\sqrt{30})=22−2\sqrt{30}.$$ But $$22−2\sqrt{30}\approx11,05...$$ 
 A: Let $f(x)=\frac{\sqrt{x+1}\ln(1+x)}{x},$ where $x>0$.
Thus, $$f'(x)=\frac{\left(\frac{\ln(1+x)}{2\sqrt{1+x}}+\frac{1}{\sqrt{1+x}}\right)x-\sqrt{1+x}\ln(1+x)}{x^2}=\frac{2x-(x+2)\ln(1+x)}{2x^2\sqrt{1+x}}\leq0$$ because
$$\left(\ln(1+x)-\frac{2x}{x+2}\right)'=\frac{x^2}{(x+1)(x+2)^2}\geq0.$$
Id est, $f$ decreases and for all $n>0$ we obtain:
$$f\left(\frac{1}{n+1}\right)>f\left(\frac{1}{n}\right)$$ or
$$\frac{\sqrt{\frac{1}{n+1}+1}\ln\left(1+\frac{1}{n+1}\right)}{\frac{1}{n+1}}>\frac{\sqrt{\frac{1}{n}+1}\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$ or
$$\sqrt{(n+1)(n+2)}\cdot\ln\frac{n+2}{n+1}>\sqrt{n(n+1)}\cdot\ln\frac{n+1}{n}$$ or
$$\left(\frac{n+2}{n+1}\right)^{\sqrt{n+2}}>\left(\frac{n+1}{n}\right)^{\sqrt{n}}.$$ 
Now, take $n=10.$
A: The question amounts to comparing $f(10)$ and $f(11)$, where $$f(x):=\sqrt{x-5}\,\log\left(1+\frac1x\right).$$
Unfortunately, this function has a maximum at about $10.4848$.
But we can use the Taylor development of the logarithm with sufficient accuracy and compare the rationals
$$5\left(\frac1{10}-\frac1{200}+\frac1{3000}-\cdots\right)^2\text{ vs. }6\left(\frac1{11}-\frac1{242}+\frac1{3993}-\cdots\right)^2.$$
Anyway, remains to determine the minimum order of the development, and computing purely by hand is more than tedious.
A: 
Theorem: Given that
  $$
u^2+v^2=2\tag1
$$
  we have
  $$
u\log(u)+v\log(v)\ge0\tag2
$$

Proof: We will show that the only critical point is when $u=v$. Assume that $u\ne v$.
By the symmetry of $(1)$ and $(2)$, we can take $u\lt v$. Constraint $(1)$ implies
$$
\frac{\mathrm{d}v}{\mathrm{d}u}=-\frac uv\tag3
$$
At any critical point, we must have
$$
\begin{align}
0
&=\frac{\mathrm{d}}{\mathrm{d}u}(u\log(u)+v\log(v))\\[3pt]
&=u\left(\frac{1+\log(u)}u-\frac{1+\log(v)}v\right)\\
&=eu\left(\frac{\log(eu)}{eu}-\frac{\log(ev)}{ev}\right)\tag4
\end{align}
$$
Since $u\lt v$, the solution to $(4)$ can be parametrized as
$$
\begin{align}
eu&=\left(1+\frac1w\right)^w\\
ev&=\left(1+\frac1w\right)^{w+1}
\end{align}\tag5
$$
Thus, $(1)$ says that
$$
\begin{align}
2e^2
&=(eu)^2+(ev)^2\\[3pt]
&=\left(1+\frac1w\right)^{2w}+\left(1+\frac1w\right)^{2w+2}\\
&=\color{#C00}{\left(\frac{w}{w+1}+\frac{w+1}{w}\right)}\color{#090}{\left(1+\frac1w\right)^{2w+1}}\\[6pt]
&\gt\color{#C00}{2}\color{#090}{e^2}\tag6
\end{align}
$$
for all finite values of $w$ because
$$
\begin{align}
\frac{w}{w+1}+\frac{w+1}{w}
&=2+\left(\sqrt{\frac{w}{w+1}}-\sqrt{\frac{w+1}{w}}\right)^2\\
&\gt2\tag7
\end{align}
$$
and Cauchy-Schwarz says
$$
\begin{align}
\left(\int_w^{w+1}x\,\mathrm{d}x\right)\left(\int_w^{w+1}\frac1x\,\mathrm{d}x\right)&\ge\left(\int_w^{w+1}1\,\mathrm{d}x\right)^2\\
\left(w+\frac12\right)\log\left(1+\frac1w\right)&\ge1\tag8
\end{align}
$$
which implies
$$
\left(1+\frac1w\right)^{2w+1}\ge e^2\tag9
$$
$(6)$ is a contradiction, which implies that the only critical point is $u=v=1$, where $u\log(u)+v\log(v)=0$. Since both $\left(0,\sqrt2\right)$ and $\left(\sqrt2,0\right)$ give $u\log(u)+v\log(v)=\frac{\log(2)}{\sqrt2}$, we have shown $(2)$.
$\large\square$

Letting $u=\sqrt{1-x}$ and $v=\sqrt{1+x}$, $(2)$ becomes
$$
\sqrt{1-x}\,\log(1-x)+\sqrt{1+x}\,\log(1+x)\ge0\tag{10}
$$
Set $x=\frac1{11}$ and we get
$$
\sqrt{\frac{10}{11}}\log\left(\frac{10}{11}\right)+\sqrt{\frac{12}{11}}\log\left(\frac{12}{11}\right)\ge0\tag{11}
$$
which gives
$$
\left(\frac{12}{11}\right)^{\sqrt6}\ge\left(\frac{11}{10}\right)^{\sqrt5}\tag{12}
$$
