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

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:
$$
(1-\lambda)^p\geq 1-\lambda\geq 1-\lambda^p.
$$
