Prove that $e^{-(1-z)\mu} \le z + (1-z)e^{-\mu}$ for $\mu > 0$ and $0 \le z < 1$. I have tried performing the power series expansion on the RHS and get
$$
z + (1-z)e^{-\mu} = 1 + (1-z)\left[-\mu + \frac{\mu^2}{2} + ...\right]
$$
And I cannot see why it is necessarily larger than $e^{-(1-z)\mu} = 1 - (1-z)\mu + (1-z)^2\mu^2/2 + ...$.
Or this inequality directly comes from some other inequality formulae?
 A: This is my first time using Jensen's inequality, so let people check this first...
Define $\alpha = e^{-\mu}.$ Since $\mu > 0,$ we have: $\ 0<\alpha<1.\ $ The original statement is thus equivalent to:

Show that
$$ \alpha^{1-z} \leq z + (1-z)\alpha\qquad \forall\ 0<\alpha<1,\ 0\leq
 z < 1. $$

Define $f(x) = \alpha^x,\ $ which we know is convex.
Jensen's inequality,
$$f(t x_1 + (1-t)x_2) \leq t f(x_1) + (1-t) f(x_2)$$
becomes:
$$ \alpha^{t x_1 + (1-t) x_2} \leq t \alpha^{x_1} + (1-t) \alpha^{x_2}. $$
Setting $x_1=0, t=z, x_2=1,$ we get the result:
$$ \alpha^{1-z} \leq z+ (1-z)\alpha.$$
$$  $$
A: Fix $z$ and let $f(y)=z+(1-z)e^{-y}-e^{-(1-z)y}$. Then $f'(y)=-(1-z)e^{-y}+(1-z)e^{-(1-z)y}=(1-z)e^{-y}(e^{yz}-1) \geq 0$. So $f$ is an increasing function with $f(0)=0$. Hence,  $f(y) \geq 0$ for all $y$ which gives the result (with $y$ replaced by $\mu$).
A: The function $t \mapsto f(t) = e^{-t}$ is convex, so that for $0 \le z \le 1$
$$
e^{-(1-z)\mu} = f(z \cdot 0 + (1-z) \cdot \mu) \le z f(0) + (1-z)f(\mu) =  z + (1-z)e^{-\mu}\, .
$$
