Application of derivatives

$\text{Consider } f(x)=x^{ \frac{1}{x}} \\ f^{'}(x) = x^{1/x} \left( \dfrac{1}{x^2} - \dfrac{\ln{x}}{x^2} \right) \\ f^{'}(x) > 0 \hspace{3mm} \forall \hspace{3mm} x<e \\ \forall \hspace{3mm} x_{1} > x_{2} \hspace{3mm} x_{1} , x_{2} > e \hspace{3mm} f(x_{1}) < f(x_{2}) \\ \text{for } x_{1}=101 \hspace{3mm} \text{ \& } \hspace{3mm} x_{2}=100 \\ 101^{\frac{1}{101}} < 100^{\frac{1}{100}}$

Most of the approaches use calculus, because that's how most people are first introduced to this idea of $x^ { \frac{1}{x} }$ . In this case, we can use the binomial theorem directly:

We want to compare $100 ^ { 101}$ with $101 ^ { 100 }$ . They look very similar, especially if we write $101^{100} = (100 + 1) ^ {100}$ . Now, let's see what we get:

$\begin{array} { l l l l l l l l l l } 101^{100} & = (100 + 1 ) ^ { 100} & = 100^{100} & + {100 \choose 1 } \times 100 ^ { 99} & + {100 \choose 2 } \times 100 ^ { 98} & + \ldots & + {100 \choose 99 } \times 100^ { 1} & + 1 \\ 100^{101} & = 100 \times 100^{100} & = 100^{100} & + 100 \times 100 ^ {99} & + 100^2 \times 100^{98} & + \ldots & + 100^{99} \times 100 \\ \end{array}$

Now, can you tell which number is larger?
Can you generalize it to all positive integers $n$ ? For which integers $n$ do we have $n^{ \frac{1}{n}} > (n+1)^{ \frac{ 1}{n+1} }$ ? (Note: Be carefull of the +1!)

Relevant wiki: Logarithms

Let $f(x) = \ln x^{1/x} = \dfrac {\ln x} x$ . Then $f'(x) = \frac{1 - \ln x}{x^2}.$ Since $\ln x > 1$ , we have $f'(x) < 0$ on the interval $x \in [100,101]$ . Thus $f(x)$ decreases on this interval; and if the logarithm decreases, so does the original function $x^{1/x}$ .

See also this similar problem .

$A =$ $\frac{1001}{100}$ $= \frac{1000}{100}$ $+ \frac{1}{100}$ $= 10 + \frac{1}{100}$

$B =$ $\frac{1011}{101}$ $= \frac{1010}{101}$ $+ \frac{1}{101}$ $= 10 + \frac{1}{101}$

$\frac{1}{100} > \frac{1}{101}$

Here is another solution that does not require calculus:

$A$ and $B$ are both positive, so a valid (in)equality between $A$ and $B$ is also valid between $A^{101}$ and $B^{101}.$ (Claim 1)

$\begin{array} {l} B^{101}=101^{(\frac {1}{101}*101)}=101\\ A^{101}=A*A^{100}=100A \end{array}$

$A^{100}=100$ and $1.01^{100} < 100^{†},$ so $A>1.01.$

Therefore, $A^{101}>(100*1.01=101=B^{101})$ and, from Claim 1, $\boxed {A>B}$

$^{†}$ If this is not intuitive, consider that $1.01x$ is the same as $x+0.01x.$

So if $x<2,$ then $1.01x<(x+0.01*2).$

Note that $1.01^{n+1}=1.01*1.01^n.$ So as long as $1.01^{n} < 2,$ $1.01^{n+1} < 1.01^n+0.02.$

Therefore, we can say that $1.01^{50}$ (think $1.01^{1+49}$ ) is less than $1.01+0.02*49=1.99.$

$1.01^{100}=(1.01^{50})^2 < (1.99)^2 < 100.$

Q.E.D.

My solution is not a proof but just an intuitive observation.

$A > B$

$100^{\frac{1}{100}} > 101^{\frac{1}{101}}$

$100^{101}>101^{100}$

$101\ln(100)>100\ln(101)$

$100\ln(100)+\ln(100)>100(\ln(100)+\epsilon)$

$100\ln(100)+\ln(100) > 100\ln(100)+100\epsilon$

$\ln(100) > 100\epsilon$

$\epsilon$ is insignificant so it's a very good guess that the inequality holds.

function f(x) = x^(1/x) reaches its max for x = e, if x tends to + infinity, it tends to 1 ! Not defined for x = 0. Has probably a saddle point somewhere e < x < + infinity.

Log(B)=log(1.01•100)/101=log(1.01)+log100/101....log(A)=log(100)/100. Log(B)/log(A)=(log(1.01)+log(100)/log100) (100/101)=(100+50 log1.01)/101. Log for base 10. So we have 3 cases log1.01=.02 or >.02or<.02 by using roots of # 10 ........ 10^.5=3.16227,10^.25=1.7782,10^.125=1.3335, 10^.0625=1.75478, 10^.03125=1.0746, 10^.015625=1.0366 so log1.01<.02. So 100+50*log1.01<101 so log(B)<log(A) so A>β######

It is simpler

$100^{1/100} ? 101^{1/101} =>$

$100^{101/100} ? 101^{101/101} =>$

$100^{1+0.01} ? 101 =>$

$100*100^{0.01} ? 101 =>$

$100^{0.01} ? 1+ 0.01 =>$

$ln(100^{0.01}) ? ln(1+0.01) =>$

$0.01*ln(100) ? 0.01+O(10^{-4}) =>$

$ln(100) ? 1 + O(0.01)$

since $ln(100)>ln(3^4)>ln(e^4)=4 =>$

$100^{1/100} > 101^{1/101}$