# The $n$-th derivative for $z$ at $z = 0$ of $F(a,b)(z)$ is no less than that of $F(\frac{a+b}{2},\frac{a+b}{2})(z)$ [closed]

The proposition the OP wants to prove is incorrect. --Aug 8, 2017

Can we find a elegant way to prove that the $n$-th derivative for $z$ at $z = 0$ of $F(a,b)(z)$ is no less than that of $F(\frac{a+b}{2},\frac{a+b}{2})(z)$ for arbitrary $n$?

Where $a,b \in \mathbb{R}^+$ and $$F(a,b)(z) = \left( \frac{1}{2} e^{(a+1)z} + \frac{1}{2}\left(\frac{1}{2}e^z+\frac{1}{2}\right)^a \right) \left( \frac{1}{2} e^{(b+1)z} + \frac{1}{2}\left(\frac{1}{2}e^z+\frac{1}{2}\right)^b \right).$$ By brute-force expanding $F$ I've made that the proposition holds where $a,b,\frac{a+b}{2} \in \mathbb{Z}$.

Furthermore, by extending the definition of $F$ to $F(a_1, a_2, \dots, a_n)$, I believe that

There will be a similar assertion with $F(a_1, \dots, a_n)$ and $F(\frac{a_1 + \dots + a_n}{n}, \dots, \frac{a_1 + \dots + a_n}{n})$.

• I'm voting to close this question because the OP seems to have answered or refuted it – Yemon Choi Aug 7 '17 at 18:18