Prove $\frac{1-z^\lambda}{1-z} < \lambda$ for $z>1$, $\lambda\in (0,1)$ This is the final part in a proof (by the definition) of the strict convexity of the exponential function, $\text{f}(x)=e^x$. It's easy to see why $\frac{1-z^\lambda}{1-z} < 1$, but proving it for a general lambda seems much more difficult.
 A: We have that by Bernoulli inequality by $y=z-1$
$$z^\lambda=(1+y)^\lambda < 1+\lambda y$$
then
$$\frac{z^\lambda-1}{z-1}<\frac{\lambda y}{y}=\lambda$$
A: I will only utilize the following well-known properties of exponentiation:1)


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*$\text{(P1)} \ $ If $a > 0$ and $m, n \in \mathbb{Z}$ with $n \neq 0$, then $a^{m/n} = \sqrt[n]{a^m}$.

*$\text{(P2)} \ $ If $a > 0$, then $s \mapsto a^s$ is continuous in $s \in \mathbb{R}$.

*$\text{(P3)} \ $ If $s > 0$, then $a \mapsto a^s$ is non-decreasing in $a \in (0, \infty)$.

*$\text{(P4)} \ $ If $a > 0$ and $s, r \in \mathbb{R}$, then $(a^r)^s = a^{rs}$.

Proof. Let $z > 1$ and $\lambda \in (0, 1)$. For any rational $r \in (0, 1)$, write $r = m/n$ for integers $0 < m < n$. Then by the AM-GM inequality,
$$ z^{r}
\stackrel{\text{(P1)}}{=} \sqrt[n]{1^{n-m} \cdot z^m}
\stackrel{\text{(AM-GM)}}{<} \frac{(n-m)\cdot 1 + m \cdot z}{n}
= 1 + r (z - 1) \tag{1} $$
Now we take limits as $r \to \lambda$ along the rationals in $\mathbb{Q}\cap(0, 1)$. Then
$$ z^{\lambda}
\stackrel{\text{(P2)}}{=} \lim_{\substack{r \to \lambda \\ r \in \mathbb{Q}\cap(0,1)}} z^{r}
\stackrel{\text{(1)}}{\leq} \lim_{\substack{r \to \lambda \\ r \in \mathbb{Q}\cap(0,1)}} (1 + r (z- 1))
= 1 + \lambda(z - 1). \tag{2} $$
Finally2), pick a rational $r \in (\lambda, 1)$ and write $\theta = \lambda / r$. Then $\theta \in (0, 1)$, and so,
$$ z^{\lambda}
\stackrel{\text{(P4)}}{=} (z^{\theta})^{r}
\stackrel{\text{(2),(P3)}}{\leq} (1 + \theta(z-1))^{r}
\stackrel{\text{(1)}}{<} 1 + r\theta (z - 1)
= 1 + \lambda(z - 1). \tag{3} $$
The desired inequality follows by rearranging $\text{(3)}$, and we are done.

1) Any sensible definition of exponentiation with positive base should entail these. Also, notice that $\text{(P3)}$–$\text{(P4)}$ can be considered as consequences of $\text{(P1)}$ and $\text{(P2)}$.
2) $\text{(2)}$ itself is not enough to establish OP's inequality. Due to the limits involved, the strictly inequality in $\text{(1)}$ needs not hold under limit, hence we need an extra argument to resort it.
A: The Bernoulli inequality is cool, but here is a more basic approach.
We want to show $\frac{1-z^\lambda}{1-z} < \lambda$, or $1-z^\lambda>\lambda-\lambda z$ (note that the inequality switches because $1-z<0$), which is the same as $1-\lambda+\lambda z-z^\lambda>0$. Fix $z$ and let $f:[0,1]\to\mathbb{R}$ be $f(\lambda)=1-\lambda+\lambda z-z^\lambda$. We see that $f(0)=0$ and $f(1)=0$. $f'(\lambda)=-1+z-z^\lambda\log(z)$. Because $z>1$, $\log(z)>0$. $f'(0)=z-1+\log(z)>0$. Note that $f'(\lambda)$ is strictly monotonically decreasing and $f$ is increasing in a neighborhood of $0$, so if $f(\lambda_0)=0$, then $f(\lambda_0+t)<0$ for all $0<t\leq 1-\lambda_0$. Thus, we cannot have any such $\lambda_0\neq1$ as $f(1)$ would be negative. Thus $f|_{(0,1)}>0$ and the inequality is true.
A: For $\lambda\in(0,1)$ and $z>1$,$$z^\lambda<z$$
or
$$z^{\lambda-1}<1.$$
Then by integration from $1$ to $z$,
$$\frac{z^\lambda-1}{\lambda}<z-1.$$
