Take a parabola $u(x)=ax^2 + bx + c$, where $a<0$, and draw a secant on an upward-sloping portion of the parabola from $(w-h, u(w-h))$ to $(w+h, u(w+h))$, for some $w \leq -\frac b{2a} - |h|$. Now find $x=\gamma$ such that $u(\gamma ) = \frac 12(u(w+h) + u(w-h))$. Prove that $\gamma < w$.
This is represented geometrically as
so can you see that it is obvious that $\gamma < w$?
However, I wanted to prove this algebraically, but I couldn't. I started off by finding the value for $\gamma $ in terms of $a$, $b$, $c$, $w$, and $h$. \begin{align*} a\gamma^2 + b\gamma + c &= \frac 12 \left(a(w+h)^2 + b(w+h) + c\right) + \frac 12 \left(a(w-h)^2 + b(w-h) + c\right)\\ &= a\left( w^2 + h^2\right) + bw + c \\ \gamma &= \frac{-b\pm\sqrt{b^2 + 4a\left(a\left(w^2 + h^2\right)+ bw\right)}}{2a} \end{align*} I tried playing with the inequalities, but I couldn't get it to work (I could only get an upper bound of $-\frac b{2a} + |h|$). I at least wanted to prove that the lower value of $\gamma$ is not correct, due to the restriction $\gamma > w - |h|$.
You already have $$ u(\gamma) = a\gamma^2 + b\gamma + c \\ = aw^2 + bw + c + ah^2\\ = u(w) + ah^2<u(w) $$ right there. Together with the fact that $u$ is increasing we get $\gamma < w$.