How do we prove that $(x-1)!\leq{(\frac{x}{2})^{x-1}}$?

Question

I plotted the graph of $y=(x-1)!$ and $y=(\frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$. Also, equality holds at $x=2$. Would someone please help me to prove it mathematically?

Answer 1

Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\\({x\over 2})^x=16$$and $$(x-1)!\not\ge ({x\over 2})^x$$

Comment

The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!\approx \sqrt {2\pi x}({x\over e})^x<({x\over 2})^x$$since $$\sqrt{2\pi x}<({e\over 2})^x$$

Answer 2

As already noted in a comment, if $x$ is intended only to have integral values $\geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, \ldots, x - 1$.

But the question is naturally read as applying to all real values of $x \geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $\Gamma(x)$.

To prove \begin{equation} \label{ineq:1}\tag{1} \log\Gamma(x) \leqslant (x - 1)\log\left(\frac{x}{2}\right) \quad (x \geqslant 2), \end{equation} because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides: \begin{equation} \label{ineq:2}\tag{2} \psi(x) \leqslant \log\left(\frac{x}{2}\right) + 1 - \frac{1}{x} \quad (x \geqslant 2), \end{equation} where $\psi$ is the Digamma function.

As a special case of the formula $$ \psi(n) = H_{n-1} - \gamma, $$ we have $$ \psi(2) = 1 - \gamma < \frac{1}{2}, $$ so \eqref{ineq:2} holds for $x = 2$.

Differentiating again, we find that it is enough to prove \begin{equation} \label{ineq:3}\tag{3} \sum_{n=0}^\infty\frac{1}{(x + n)^2} \leqslant \frac{1}{x} + \frac{1}{x^2} \quad (x \geqslant 2). \end{equation} Indeed, $$ \sum_{n=1}^\infty\frac{1}{(x + n)^2} < \sum_{n=1}^\infty\frac{1}{(x + n - 1)(x + n)} = \sum_{n=1}^\infty\left(\frac{1}{x + n - 1} - \frac{1}{x + n}\right) = \frac{1}{x}. $$ This proves \eqref{ineq:3}, therefore \eqref{ineq:2}, and therefore \eqref{ineq:1}, with strict inequality for $x > 2$.