How do you prove Gautschi's inequality for the gamma function?

The strict log-convexity of $\Gamma$ (see the end of this answer) implies that for $0< s <1$, $$ \Gamma(x+s)<\Gamma(x)^{1-s}\Gamma(x+1)^s=x^{s-1}\Gamma(x+1)\tag{1} $$ which yields $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}\tag{2} $$ Again by the strict log-convexity of $\Gamma$, $$ \Gamma(x+1)<\Gamma(x+s)^s\Gamma(x+s+1)^{1-s}=(x+s)^{1-s}\Gamma(x+s)\tag{3} $$ which yields $$ \frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+s)^{1-s}<(x+1)^{1-s}\tag{4} $$ Combining $(2)$ and $(4)$ yields $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}\tag{5} $$

I'll probably leave this standing for two days before posting a summary of Gautschi's paper.

Here is the long-overdue follow-through. I have slightly changed a few notations, but this is otherwise Gautschi's original argument.

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What Gautschi actually proves in his paper is the more general inequality

$$\exp((s-1)\psi(n+1))\le\frac{\Gamma(n+s)}{\Gamma(n+1)}\le n^{s-1},\; 0\le s\le1,n\in\mathbb Z^{+}\tag{1}\label{1}$$

where $\psi(n)$ is the digamma function.

Gautschi considers the function

$$f(s)=\frac1{1-s}\log\left(\frac{\Gamma(n+s)}{\Gamma(n+1)}\right)$$

over $0\le s <1$, from which we have $f(0)=\log(1/n)$ and

$$\lim_{s\to 1}f(s)=-\psi(n+1)$$

via l'Hôpital. Then we have

$$(1-s)f'(s)=f(s)+\psi(n+s)$$

and then by letting

$$\varphi(s)=(1-s)(f(s)+\psi(n+s))$$

we have $\varphi(0)=\psi(n)-\log n<0$, $\varphi(1)=0$, and $\varphi'(s)=(1-s)\,\psi ^{(1)}(n+s)$ (where $\psi ^{(1)}(n)$ is the trigamma function).

Now, since $\psi ^{(1)}(n+s)=\psi ^{(1)}(s)-\sum\limits_{k=0}^{n-1}\frac1{(s+k)^2}$ is always positive, we have that $\varphi(s)<0$, from which we deduce that $f(s)$ is monotonically decreasing over $0<s<1$ (i.e., $f'(s)<0$). Therefore

$$-\psi(n+1)\le f(s)\le\log\frac1{n}$$

which is equivalent to $\eqref{1}$. The inequality in the OP can then be deduced from the inequality $\psi(n)<\log n$.