The inequality $$x^s\le 1+ s(x-1)$$ for $x>0$ and any $0<s<1$ is usually proven firstly for rational $s$ using AM-GM, and then by some density of the rationals argument. My question:
Are there any mistakes in the following alternative derivation?
For simplicity let's consider the case $s=1/k$, where $k$ is a natural number.
Then the inequality we want to prove, $x^{1/k}<1+(x-1)/k$, is equivalent to $$ x \le (1+\frac{x-1}{k})^k$$
By the binomial theorem, the RHS is equal to
$$\sum_{i=0}^{k}\binom{k}{i}(\frac{x-1}{k})^i=1+(x-1)+\sum_{i=2}^{k}\binom{k}{i}(\frac{x-1}{k})^i$$
Plugging this into the former inequality reduces it to proving that
$$\sum_{i=2}^{k}\binom{k}{i}(\frac{x-1}{k})^i\ge0$$
This is obvious for $x\ge1$, so let's consider $0<x<1$. We notice that in this case, when $i$ is even the corrisponding term in the above sum is positive, and when $i$ is odd the term is negative. So if we prove that the sum of two consecutive terms, that is
$$\binom{k}{i}(\frac{x-1}{k})^i+\binom{k}{i+1}(\frac{x-1}{k})^{i+1}\ge0$$
for a certain even $2\le i\le k$, then the sum of every consecutive two terms in the previous sum is positive and so the sum will be positive too.
Simplifying the binomials in the former expression we get:
$$(i+1)(\frac{x-1}{k})^i+(k-i)(\frac{x-1}{k})^{i+1}\ge0$$
and simplifying the other terms (excluding the case $x=1$), remembering that $x-1<0$, we end up with:
$$(i+1)-(k-i)(\frac{1-x}{k})\ge0$$
which is equivalent to
$$(i+1)\ge(1-i/k)(1-x)$$
On the RHS we have a product of two numbers both less than $1$, and on the LHS an integer $\ge 3$, so the inequality is obvious.