parmenides51 wrote: For each real number $p > 1$, find the minimum possible value of the sum $x+y$, where the numbers $x$ and $y$ satisfy the equation $(x+\sqrt{1+x^2})(y+\sqrt{1+y^2}) = p$. Write $x=\frac 12\left(u-\frac 1u\right)$ and $y=\frac 12\left(v-\frac 1v\right)$ with $u,v>0$ Equation becomes $uv=p$ $x+y=\frac 12\left(u+v-\frac 1u-\frac 1v\right)$ $=\frac{u+v}2\left(1-\frac 1p\right)$ And so $x+y=\frac{u+\frac pu}2\left(1-\frac 1p\right)$ And minimum is trivially reached when $u=\sqrt p$ and is $\boxed{\frac{p-1}{\sqrt p}}$