Is is true that $(1-\lambda)^p\leq1-\lambda^p$ for all $\lambda\in [0,1]$ and $p\in (0,1)$?

Question

For all $\lambda\in\Bbb{R}$, $0\leq\lambda\leq1$ and $0<p<1$, $p\in\Bbb{R}$ is the following true? $$(1-\lambda)^p\leq1-\lambda^p$$

Answer 1

$$(1-\lambda)^p > 1-\lambda$$ and $$\lambda^p > \lambda$$

so their sum is bigger than $1$

Answer 2

Actually, the reverse inequality is true. Using the concavity of $f(x)=x^p$, we have $$ \lambda^p=f(\lambda\times 1+(1-\lambda)\times0)\geq \lambda f(1)+(1-\lambda)f(0)=\lambda. $$ Similarly, $$ (1-\lambda)^p\geq 1-\lambda. $$ Putting these together: $$ (1-\lambda)^p\geq 1-\lambda\geq 1-\lambda^p. $$