# Why $\texttt{Sqrt}[\textrm{positive value}] > 0$ sometimes simplifies to true, sometimes not

Why this does not evaluate to True:

Assuming[-2 bRis bRm2 + (aRx rho + vR0)^2 > 0, Simplify[Sqrt[-2 bRis bRm2 +
(aRx rho + vR0)^2] > 0, reals]]
(* Sqrt[-2 bRis bRm2 + (aRx rho + vR0)^2] > 0 *)


if this does:

Assuming[-2 bRis bRm2 > 0, Simplify[Sqrt[-2 bRis bRm2] > 0]]
(* True *)


The second case would imply that Sqrt uses only the principal root.

• Instead of reals, use Reals. Aug 10, 2023 at 15:32

I would also use Reduce. It tries harder than Simplify or FullSimplify.

For instance:

Assuming[-2 bRis bRm2 + (aRx rho + vR0)^2 > 0,
Simplify@
Reduce[Implies[$Assumptions, Sqrt[-2 bRis bRm2 + (aRx rho + vR0)^2] > 0]] ] (* True *)  Or simply: Assuming[-2 bRis bRm2 + (aRx rho + vR0)^2 > 0, Reduce[Implies[$Assumptions,
Sqrt[-2 bRis bRm2 + (aRx rho + vR0)^2] > 0], Reals]
]

(*  True  *)


Or equivalently:

Reduce[
Implies[
-2 bRis bRm2 + (aRx rho + vR0)^2 > 0,
Sqrt[-2 bRis bRm2 + (aRx rho + vR0)^2] > 0
],
Reals]

(*  True  *)


If you solve inequality Reduce is better:

Reduce[Sqrt[-2 b1 b2 + (a r + v)^2] > 0 && -2 b1 b2 + (a r + v)^2 >
0 && a > 0 && b1 > 0 && b2 > 0 && r > 0 && v > 0]

(*  v > 0 && r > 0 && a > 0 && b2 > 0 &&
0 < b1 < (a^2 r^2 + 2 a r v + v^2)/(2 b2)  *)


If you use Reals insted of what I did, you get an answer, but it will be considerably longer. That's why I did not do it.

Have fun!