Vtyjkyuk

Let $f(x)=(1+x)^frac1xleft(1+frac1xright)^x, 0<xleq 1.$
Prove that $f$ is strictly increasing and $e<f(x)leq 4.$

In order to study the Monotonicity of $f$, let
$$g(x)=log f(x)=frac1xlog (1+x)+xlog left(1+frac1xright).$$
And $f$ and $g$ has the same Monotonicity. By computation,
$$g'(x)=frac1x^2left(fracx1+x-log (1+x)right)+log left(1+frac1xright)-frac11+x.$$
As we know $fracx1+x-log (1+x)leq 0$ and $log left(1+frac1xright)-frac11+xgeq 0$. So it does not determine the sign of $g'(x)$.
If we compute the second derivative $g''(x)$, you will find it is also difficult to determine the sign of $g''(x)$.
Our main goal is to prove $$frac1x^2left(fracx1+x-log (1+x)right)+log left(1+frac1xright)-frac11+x>0.$$

Is there some tricks to prove this result. Any help and hint will welcome.

The right inequality.

We can use the TL method here.

We need to prove that $$(1+a)^b(1+b)^aleq4$$
for $a>0$, $b>0$ such that $ab=1$, which is $$fracln(1+a)a+fracln(1+b)bleq2ln2.$$
But $$sum_cycleft(ln2-fracln(1+a)aright)=sum_cycf(a),$$
where $f(a)=ln2-fracln(1+a)a-(ln2-0.5)ln a$.

Easy to show that $f(a)geq0$ for all $0<aleq11$.

But for $a>11$ we obtain: $$(1+a)^b(1+b)^a=(1+a)^frac1aleft(1+frac1aright)^a<(1+11)^frac111e<4.$$

The left inequality.

We need to prove that:
$$frac1xln(1+x)+xlnleft(1+frac1xright)>1$$ or $g(x)>0$, where
$$g(x)=(1+x^2)ln(1+x)-x^2lnx-x.$$
Indeed, $$g'(x)=2xln(1+x)+frac1+x^21+x-2xlnx-x-1=2xleft(lnleft(1+frac1xright)-frac11+xright)>0$$ because
$$left(lnleft(1+frac1xright)-frac11+xright)'=-frac1x(1+x)^2<0$$ and
$$lnleft(1+frac11right)-frac11+1>0.$$
Id est, $$g(x)>lim_xrightarrow0^+g(x)=0$$ and we proved that $e<f(x)leq4.$

Now, we'll prove that $f$ increases on $(0,1].$

By your work we need to prove that
$$frac1x^2left(fracx1+x-ln(1+x)right)+lnleft(1+frac1xright)-frac11+xgeq0$$ for all $0<xleq1$ or
$$(x^2-1)ln(1+x)-x^2lnx+frac(1-x)x1+xgeq0$$ and since $$lnxleqfrac2(x-1)1+x,$$ it's enough to prove that
$$-(1-x^2)ln(1+x)+frac2x^2(1-x)1+x+frac(1-x)x1+xgeq0$$ or
$$ln(1+x)leqfracx(2x+1)(1+x)^2,$$ which is smooth.

Michael Rozenberg's answer has it all.

Here are two remaining proofs in Michael Rozenberg's answer:

1. Show: $lnxleqfrac2(x-1)1+x$

We have $ln(1+y)leqfrac2y2+y$ for $-1<y<0$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities),
i.e. $ln(x)leqfrac2(x-1)1+x$ for $0<x<1$. This is exactly what needs to be shown.

2. Show: $ln(1+x)leqfracx(2x+1)(1+x)^2$

We have $ln(1+x)leqfracxsqrt1+x$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), so we may prove

$frac1sqrt1+xleqfrac2x+1(1+x)^2$ or $(1+x)^3-(1+2x)^2leq 0$ or $x^2 - x - 1<0$ or $x( 1-x) + 1>0$ which is true since $1-x ge 0$.

Using
$$dfracxx+1<xlnleft(1+dfrac1xright)<1$$
and for $xtodfrac1x$
$$dfrac1x+1<dfrac1xlnleft(1+xright)<1$$
then
$$dfrac1xlnleft(1+xright)+xlnleft(1+dfrac1xright)>1$$