How to prove this inequality? What method should I choose? $e^{1/x} < 27/x\beta$, for all integer $x>2$
I do by calculator that this is indeed true. But how to prove this in general?
 A: Take log of both sides to get $$\log x+\log (x-1)+\log(x-2)<(x+1)\log 2$$
Bound all $\log$ terms from above by a tangent at $x_0$ using $\log x\leq \log(x_0)+\frac 1 {x_0}(x-x_0)$ to get
$$RHS<3\log(x_0)+\frac 3 {x_0}(x-x_0)-\frac 3 {x_0}<(x+1)\log 2$$
Choose $x_0=\frac 3 {\log 2}\approx 4.328$ to match linear terms on both sides. We are left to verify that
$$3\log \frac 3{\log 2}<3+2\log 2$$
Turns out that $E=LHS-RHS\approx 0.009$ (close, but no cigar) so we need to include quadratic terms in our approximation:
$$\log x\leq\log(x_0)+\frac 1 {x_0}(x-x_0)-\frac 1 {2(\max\{x,x_0\})^2}(x-x_0)^2$$
Summing 3 terms and simplifying we are left to show that
$$E\leq\frac {3(x-x_0-1)^2+2}{2(\max\{x,x_0\})^2}$$
Since $E<1/100$ for this not to hold, $\max\{x,x_0\}>10$ but in that region the first term in the numerator will be comparable (compared to smallness of $E$) to the denominator as $x_0<5$.
A: If $x$ is an integer, you can show it in the following way (I use $n$ instead of $x$:
$$\left(\frac{n(n-1)(n-2)}{2}\right)^{1/n} < 2 \tag{0}$$
Take the n-th power and get
$$\frac{n(n-1)(n-2)}{2}< 2^n \tag{1}$$
Note that 
$$\frac{n(n-1)(n-2)}{2} = 3{n \choose 3}  $$ 
and so the above statements is a consequence of the binomial theorem:
$$3{n \choose 3} \le (n-5){n \choose 3} = \sum_{k=3}^{n-3}{n \choose 3} \lt \sum_{k=3}^{n-3}{n \choose k} \lt \sum_{k=0}^{n}{n \choose k} =2^n $$
if $n \ge 8$. ${n \choose 3} \le {n \choose k}$ for $ 3 \lt k \lt n-3$ is a property of the pascal triangle.
But the inequation ($0)$can be proofed by induction, too.
Set 
$$p(n)=\frac{n(n-1)(n-2)}{2}$$
Then you have to show 
$$p(n)<2^n$$
But we have
$$p(n+1)=\frac{p(n+1)}{p(n)}p(n) \lt 2 \cdot 2^n = 2^{n+1} \tag{2}$$
if $n \ge 8$, because
$$\frac{p(n+1)}{p(n)}=\frac{n+1}{n-2}=1+\frac{3}{n-2}\le 2$$
For $n \in {1,2,3,4,5}$ you can check the inequation directly by plugging in these numbers. Together with $(2)$ this shows that the statement is valid for all $n \in \mathbb{N}$.
A: Consider the function over real values of $x$, $$f(x) = 2^x - \tfrac12 x(x-1)(x-2)$$
Find expressions for $f'(x)$, $f''(x)$, and $f'''(x)$.
By trial and error, find an integer value of $x$ (no less than $3$) for which
$f'(x)$, $f''(x)$, and $f'''(x)$ are all positive.
You can guess, or you can try $x=3$, then $x=4$, and so forth
until you find a suitable $x$. 
(I have verified that you don't have to search very far.)
For the rest of this answer, I will use $x_0$ to denote the value of $x$
you found in the previous step. Of course when you do this, having actually
found $x_0$ you will simply use the numeric value you found.
Since $f'''(x)$ is an increasing function of $x$ everywhere
(if this is not sufficiently obvious, take its derivative one
more time and notice that the result is always positive),
$f'''(x) > 0$ whenever $x \geq x_0$.
Since $f'''(x) > 0$ whenever $x \geq x_0$ and $f''(x_0) > 0$,
$f''(x) > 0$ whenever $x \geq x_0$.
Since $f''(x) > 0$ whenever $x \geq x_0$ and $f'(x_0) > 0$,
$f'(x) > 0$ whenever $x \geq x_0$.
Since $f'(x) > 0$ whenever $x \geq x_0$ and $f(x_0) > 0$,
$f(x) > 0$ whenever $x \geq x_0$.
You can now calculate the values of $f(x)$ for $x=3$, $x=4$, and so forth
up to and including $x=x_0 - 1$, to a sufficient precision to show that
all are positive.
(In fact, you may already have done this in the "trial and error" step.)
You will then have shown that
$f(x) > 0$ for all integers $x$ such that $x \geq 3$, that is, 
for all integers $x > 2$.
You can show (using logarithms or other methods) 
that if $x > 2$ and $f(x) > 0$, then
$$\left(\frac{x(x-1)(x-2)}{2}\right)^{1/x} < 2.$$
But since $f(x) > 0$ for all integers $x$ such that $x > 2$,
as you will already have shown,
it is also true that $\left(\frac{x(x-1)(x-2)}{2}\right)^{1/x} < 2$
for all integers $x$ such that $x > 2$.
Alternatively, you could let 
$g(x) = 2 - \left(\frac{x(x-1)(x-2)}{2}\right)^{1/x}$
and do the same reasoning as above the the derivatives
$g'(x)$, $g''(x)$, and $g'''(x)$, but I think the formulas for
those derivatives would be much more difficult to work with.
