Any Simple Way to Prove Black Scholes Type Identies?

A certain complicated option pricing formula results in products of Black Scholes $N$ components like this:

$-p_1N(d_1)N(d_6)+p_sN(d_2)N(d_5)>?0$ where $p_s>p_1$

Trying to find a simple way to prove if this identity is true or not without having to use differential geometry on a multi variable function. Taylor series don't work because the Ln's and $a*t^{1/2}$ part can be very big. interest set to zero for simplicity

\begin{align} d_1 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{p_1}{p_s}\right) + \left(0 + \frac{\alpha^2}{2}\right)t\right] \\ d_2 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{p_1}{p_s}\right) + \left(0 - \frac{\alpha^2}{2}\right)t\right] \\ d_5 &= \frac{1}{\alpha\sqrt{t}}\left[ \left(0 + \frac{\alpha^2}{2}\right)t\right] \\ d_6 &= \frac{1}{\alpha\sqrt{t}}\left[\left(0 - \frac{\alpha^2}{2}\right)t\right] \\ \end{align}

Trying to prove this with differentials requires taking the derivative of a complicated function with respectto the time/vol variable and then the p_s an p_1 variables to determine if the manifold has a maximum above zero

At the endpoints $at \to 0$ and $at \to infinity$ it equals zero, so the function is a frown or a smile that has positive or negative curvature. This is an extremely hard problem because linear approximations don;t work

• $N(d_6)=1-N(d_5)$ so we have $-bN(d_1)+bN(d_5)N(d_1)+p_sN(d_2)N(d_5)$ – user151781 Nov 26 '14 at 2:33
• Consider a stronger bound for this problem. We know $N(d_5)/N(d_1)>1$ Divide entire expression by $N(d_1)$ to get a stronger bound and we have $-b+bN(d_5)+p_sN(d_2)$ greater or equal to $0$ . This also obeys the endpoint conditions for $at$. This smaller expression is easier to handle with calculus – user151781 Nov 26 '14 at 3:05