How can one prove that
$$\sqrt[105]{105}>\sqrt[106]{106}\text{ ?}$$ Induction on the statement $$\sqrt[n]{n}>\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n>2$$ would yield $\sqrt[3]{3}>\sqrt[4]{4}$ at the base step $n=3$, which we cannot assume.
So, based on the properties of powers and square roots alone, can we prove the first statement?
EDIT: No calculus, no functions, not even logs.
It's more of a riddle that anything else.
On
Graphing $y=x^{\frac{1}{x}}$ , Y has a maximum value at x = e.
It's all asymptotic to 1 as x approaches infinity
Updated with graph to show exact area of interest. When properly scaled, it's clear that Y is dropping like a rock as X increases. The good answers here still don't adhere to Op's request for "powers and square roots alone" and his edit of "No calculus, no functions, not even logs." The highest rated answer? ln x from the start.
On
This is equivalent to: $$105^{1/105}>1+\frac1{105}$$ (which apparently can be proven using binomial expansions).
Proof: \begin{align} 105^{1/105}&>1+\frac1{105}\\ 105^{1/105}&>\frac{106}{105}\\ 105^{1+1/105}&>106\\ 105^{106/105}&>106\\ 105^{1/105}&>106^{1/106} \end{align}
On
This generalizes to $n^{n+1}>(n+1)^n$ or, by the binomial theorem, $$n.n^n>n^n+{n\choose1}n^{n-1}+{n\choose2}n^{n-2}+\cdots+{n\choose {n-1}}n+1.$$ Noting that the first two terms are exactly $n^n$, it remains to show that each term, starting from the 3rd -- Thanks @TonyK, on the RHS is less than $n^n$, which is clear for $n$ large (Hint: What is $n\choose k$?).
One may then ask, but the RHS has $n+1$ terms. The answer is the last term is $1$, so one may combine last two terms and compare to $n^n$.
Also note that the first two terms are exactly $n^n$. That explains why we needs $n\ge3$.
On
All you need here is the binomial theorem for positive integer exponents:
$$(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$$
Fix $n \ge 3$, and put $x = \frac{1}{n}$:
$$\left(1+\frac{1}{n}\right)^n = \sum_{r=0}^n \binom{n}{r}n^{-r}$$
Now, $\binom{n}{r} \le n^r$ for all $r=0,\ldots,n$ (easily proven by induction; or just look at the expression $\binom{n}{r} = \frac{n}{r}\frac{n-1}{r-1}\cdots\frac{n-r+1}{1}$). So
$$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^n n^rn^{-r} = n+1$$
This is not quite good enough for our purposes, but we can improve the estimate by considering the last two terms separately:
$$\begin{align} \left(1+\frac{1}{n}\right)^n &\le \sum_{r=0}^{n-2} n^rn^{-r} + \binom{n}{n-1}n^{-(n-1)}+\binom{n}{n}n^{-n} \\ &= n - 1 + n^{-(n-2)}+n^{-n} \\ &< n - 1 + 2n^{-(n-2)} \\ &< n \end{align}$$
because $n^{n-2} > 2$ if $n \ge 3$, so $n^{-(n-2)} < \frac12$.
Therefore $n > (1+\frac{1}{n})^n$ for all $n \ge 3$. Multiplying both sides by $n^n$ gives $n^{n+1} > (n+1)^n$. Taking the $n(n+1)$-th root gives, for all $n \ge 3$, $$\sqrt[n]{n} > \sqrt[n+1]{n+1}$$
On
Note that $$1+\frac{1}{n}<e^{1/n}$$
So $$\left(1+\frac{1}{n}\right)^n < e$$
For $n>e$ then:
$$\left(\frac{n+1}{n}\right)^n<e<n$$
Multiply both sides by $n^n$ and you get:
$$(n+1)^n<n^{n+1}$$
That's not an induction proof, however.
If you don't know the power series for of $e^{x}$, you might prove directly that:
$$\left(1+\frac{1}{n}\right)^n <3.$$ Namely:
$$\left(1+\frac{1}{n}\right)^n =\sum_{k=0}^n \binom{n}{k}\frac{1}{n^{k}}<\sum_{k=0}^\infty \frac{1}{k!}$$
And you can easily see that for $k\geq 1, \frac{1}{k!}\leq\frac{1}{2^{k-1}}$ so the sum is less than $3$.
One way to do this is to simply note that the function $x^{1/x}$ is decreasing over a range including $105$ and $106$.
To this end, it suffices to show that $\ln[x^{1/x}] = \frac{\ln(x)}{x}$ is decreasing over such a range.
We calculate: $$ \frac{d}{dx}\left( \frac{\ln(x)}{x} \right) = \frac{1 - \ln(x)}{x^2} $$ So, this function is decreasing when $1 - \ln(x) < 0 \implies x > e$. So, the function is decreasing over the range $(e,\infty)$, which suffices for our purposes.
For that base step, should you want to prove this without calculus: $$ 3^{1/3} > 4^{1/4} \iff\\ 3^4 > 4^3 \iff \\ 81 > 64 \quad \checkmark $$