Showing $\left(\sqrt{1-p}-\sqrt{1-q}\right)^2+\left(\sqrt{p}-\sqrt{q}\right)^2\leq \left(\frac{1}{p}+\frac{1}{1-p}\right) (p-q)^2 $ for $0

Question

I am trying to show that

$$ \left(\sqrt{1-p}-\sqrt{1-q}\right)^2+\left(\sqrt{p}-\sqrt{q}\right)^2\leq \left(\frac{1}{p}+\frac{1}{1-p}\right) (p-q)^2 $$ for $0<p<1, 0<q<1$.

If you check this in Mathematica, it simply says this is True. But I cannot think any well-known inequality to apply here.

Answer 1

$(\sqrt {1-p} -\sqrt {1-q})^{2}=\frac {(p-q)^{2}} {(\sqrt {(1-p)}+\sqrt {1-q})^{2}}\leq \frac {(p-q)^{2}} {1-p}$. Similarly the second term does not exceed $\frac {(p-q)^{2}} p$. Just add these two inequalitities.