Looking for an other method to compare $100!$ and $50^{100}$ There is a question asked me to compare $100! \ and \ 50^{100}$. I think more than 2 hours to solve it.
I show my work below, but I looking for other Ideas to prove the inequality.  Thanks in advance for any hints or new Ideas.
my work :
$$50 \times 50 >  1 \times 99 \times 2\\
50 \times 50 > \ 2 \times 98\\
50 \times 50 > \ 3 \times 97\\
50 \times 50 > \ 4 \times 96\\ \vdots \\
50 \times 50 > \ 48 \times 52=(50+2)(50-2)=50^2-4\\
50 \times 50 > \ 49 \times 51=(50+1)(50-1)=50^2-1\\
50 \times 50 \geq 50 \times 50 $$ then multiply them
$$50^{49}\times 50^{49}\times 50^2 >1.2.3...50....99.(2.50)\\50^{100}>100!$$
Remark: I use numerical approximation for $50^{100}\sim 7.8886\times 10^{169} $ and $100!=9.33236\times 10^{157}$ But is not interesting( Ithink).
 A: Well they both have $100$ factor components.
The LHS the factor components run from $1$ to $100$.  The RHS the factors are a constant coomponent of $50$.
$50$ is almost the arithmetic average of $1$ to $100$.  By the A.M. G.M inequality that is greater than the geometric mean of $1$ to $100$ which is  $\sqrt[100]{1*2*3*....*100}$.
And so we ought to have $\sqrt[100]{100!} < 50$ and the $100! < 50^{100}$.
.... but the arithmetic average is actually $50.5$ so we have $\sqrt[100]{100!} < 50.1$.
But we can probably tweak it not to make such a difference.
.....
And indeed you did tweak it!
$2\times 4 \times ....... \times 98 =98!$ and $\sqrt[97]{98!}$ is the geometry mean of $2,....,98$.  And the arithmetic mean of $2,...., 98$ is $50$.
So we have $\sqrt[97]{98!} < 50$ and so $98! < 50^{97}$
Now the tweak: $99\times 100 < 100^2 = 4\times 50^2 < 50^3$ so $100! = 98!\cdot (99\times 100) < 50^{97} \times 50^3= 50^{100}$
The tweak makes sense to look at the edge the difference between geometric mean and arithmetic mean is greatest when the numbers are far apart.  In this case we managed to be a product of three terms which wasn't just less than the cube of the AM.  It was barely significantly more than the square of the AM.
A: Definitely quicker than 2 hours, plus one more method to have in arsenal (similar to this, by the way). At some point I will use
$$\log(1+x)\leq x, \forall x \geq 0$$
$$\log{2}<\frac{3}{4} \iff 16 < e^3$$
(in fact $2.6^3>16 \iff 13^3 > 5^3\cdot 4^2=5\cdot400$)
and $$\log{101}<\log{128}=7\log{2}<\frac{21}{4}$$
Let's look at $\sum\limits_{k=1}^{100}\log{k}$. Function $f(x)=\log{x}$ is monotone ascending, thus:
$$\sum\limits_{k=1}^{100}\log{k}=
\sum\limits_{k=1}^{100}\log{k} \cdot (k+1 - k)\leq \int\limits_{1}^{101}\log{x} dx =
x (\log{x}-1) \Big|_{1}^{101}=\\
101\log{101}-101+1=
101\log{101}-100=\\
100\log{100}+100\log{\left(\frac{101}{100}\right)}+\log{101}-100<\\
100\log{100}+100\log{\left(1+\frac{1}{100}\right)}+\frac{21}{4}-100\leq\\
100\log{100}+1+\frac{21}{4}-100=\\
100\log{50}+100\log{2}+\frac{25}{4}-100<\\
100\log{50}+100\cdot \frac{3}{4}+\frac{25}{4}-100<100\log{50}$$
A: $$\ln (100!)=\sum_{n=1}^{100}\ln n<\sum_{n=1}^{100}\int_n^{n+1}\ln x\,dx=$$ $$=\int_1^{101}\ln x\,dx=$$ $$=[101\ln 101 - 101]-[1\ln 1 -1]=$$ $$=101\ln 101 - 100.$$ Therefore $$\ln (100!)-\ln (50^{100})<101\ln 101 -100-100\ln 50=$$ $$=(100)(1.01\ln 100  +1.01\ln 1.01-1-(\ln 100-\ln 2))=$$ $$=(100)(0.01\ln 100 +1.01\ln 1.01+\ln 2 -1)<0$$ because $\ln 2<0.70,$ and $\ln 100<6,$ and $1.01\ln 1.01<(1.01)(0.01)<0.02.$
