I want to prove that it holds for $p,q \in (0,1)$ and $\gamma \in (-\infty, 0)$ $$ p^{(1-\gamma)} q^\gamma + (1-p)^{(1-\gamma)} (1-q)^\gamma \geq 1. $$
I proved it by regarding the left-hand side of the inequality as a function in $p$ for $q$ fixed, taking the derivative and using some more (rather technical) arguments to show that this function is minimal for $p=q$ with function value $1$.
That is, fix $q\in (0,1)$ and define a function $h \colon (0,1) \to \mathbb{R}$ by
$$ p \mapsto p^{(1-\gamma)} q^\gamma + (1-p)^{(1-\gamma)} (1-q)^\gamma. $$
Then $$ h^\prime(p) = \dfrac{q^{\gamma}\cdot\left(1-{\gamma}\right)}{p^{\gamma}}-\dfrac{\left(1-q\right)^{\gamma}\left(1-{\gamma}\right)}{\left(1-p\right)^{\gamma}} $$ and $h^\prime(p)=0$ iif $p=q$. (One can verify that this is the minimizer indeed.) In that case, $h(q)=1$ and this yields the claim.
Are there less technical and more direct arguments/ways to prove the inequality? Thank you. :)
We can use Bernoulli inequality: $(1+ x)^r \ge 1 + rx$ for all $r < 0$ and $x > -1$.
The desired inequality is written as $$p (q/p)^r + (1-p)((1-q)/(1-p))^r \ge 1.$$
For convenience, we replace $\gamma$ with $r$. By Bernoulli inequality, we have $$\mathrm{LHS} \ge p[1 + (q/p - 1)r] + (1-p)[1 + ((1-q)/(1-p) - 1)r] = 1.$$
We are done.