Proof that $1 \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e}$ with $0Is there a way to prove that 
$$
1 \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e} \\
\leftrightarrow \sqrt{1-p} \cdot \sqrt{1-p}+\sqrt{p} \cdot \sqrt{p} \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e}
$$
with $0<p<1$ and $0<e<p$
I have been struggling on this all day, and I thought that one of you might have the answer? It may have to do with the concavity of the square-root function? 

Let $p=u^2$ and $p-e=v^2$, with $0\lt v\lt u\lt1$.  The inequality becomes
$$1\ge\sqrt{1-u^2}\sqrt{1-v^2}+uv$$
or
$$1-uv\ge\sqrt{1-u^2}\sqrt{1-v^2}$$
Since $1-uv\gt0$, this inequality is equivalent to
$$(1-uv)^2\ge(1-u^2)(1-v^2)$$
which simplifies to $u^2-2uv+v^2\ge0$, which holds since $u^2-2uv+v^2=(u-v)^2$.
