suppose $f(x) = -x^2 + bx + 1$ and and $g(x) = x^2 + 2x + c$ are such that
$max \quad f(x)\le min\quad g(x)$ as $x$ varies over the set of real numbers. The least possible value of $c$
what I tried is to the maximum value of $f(x)$ and minimum value of $g(x)$ and apply the given inequality. But I am getting an equation in $c$ and $b$ and I don't know how to proceed further.
Observe that $g(x) = x^{2} + 2x + c = (x^{2} + 2x + 1) + c - 1 = (x+1)^{2} + c - 1\geq c - 1$. Moreover, $f(x) = -x^{2} + bx + 1 = (-x^{2} + bx - b^{2}/4) + b^{2}/4 + 1 = -(x-b/2)^{2} + b^{2}/4 + 1 \leq b^{2}/4 + 1$. Hence we may conclude that $c$ must satisfy the following inequality: \begin{align*} \max f(x) = b^{2}/4 + 1 \leq c - 1 = \min g(x) \Leftrightarrow c\geq b^{2}/4 + 2\Leftrightarrow c_{min} = b^{2}/4 + 2 \end{align*} Hope this helps.
Since $f''(x)<2$ thus $f(x)$ at critical point is always a local maxima
$$f'(x)=-2x+b\tag{x=b/2 is critical point}$$
Similarly, on the same lines, $g(x)$ at critical point is always a local minima $$g'(x)=2x+2\tag{x=-1 a crital point}$$
$$f(\frac{b}{2})\le g(-1)$$ $$-\frac{b^2}{4}+\frac{b^2}{2}+1 \le -1+c$$
Therefore $$c \ge\frac{b^2}{4}+2$$
