Proof that $1 \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e}$ with $0<p<1$ and $0<e<p$

Question

Is 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?

Answer 1

after squaring two times and simplifications we get $$e^2>0$$ which is true.

Answer 2

Try using Cauchy-Schwarz, see what you get from there. (Which is a result based on concavity, actually!)

Answer 3

By C-S $$\sqrt{(1-p)(1-p+e)}+\sqrt{p(p-e)}\leq\sqrt{(1-p+p)(1-p+e+p-e)}=1$$

Answer 4

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 $u^2-2uv+v^2=(u-v)^2$.