Let us suppose that $F(x)=ax^{2}+bx+c$ and $G(x)=Ax^2+Bx+C$ where $a, b, c, A, B, $ and $C$ are real numbers with $a > 0$ and $A > 0$.
Let us suppose that the minimum value of the quadratic function defined by $F(x)$ is greater than or equal to the minimum value of the quadratic function defined by $G(x)$. Does it necessarily hold that $G(x) \leq F(x)$ for every real number $x$?
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For any quadratic, we can use a change of variables to set $F(x)=x^2$ and $G(x)=m(x-b_1)(x-b_2)$ with $m > 0$. Since $G(x)$ has a vertex $\leq$ that of $F(x)$ at $(0,0)$ it has two real roots $b_1,b_2$. It will be convenient not to deal with separate $b_i$ terms so let's set $b_2=kb_1$. In this way, if $k=1$ the roots are identical, if $k=-1$ the roots are symmetric and the vertex of $G(x)$ is directly below that of $F(x)$.
We examine if parabolas intersect by setting $x^2=m(x-b)(x-kb)$, leading to the quadratic:
$$x^2(m-1)-xmb(k+1)+mkb^2=0$$
Case 1: $m=1$
The $x^2$ terms cancel, and $x=\frac{bk}{k+1}$ gives the intersection of the parabolas. Only if $k=-1$ does no $x$ satisfy, which is when the parabola of $G(x)$ is shifted directly down from $F(x)$ by a constant. This would be $G(x)=x^2-c$. Intuitively, if we have identical parabolas $(m=1)$, the only way to prevent intersection is to shift one vertically, in this case $G(x)$ down.
Case 2: $m\neq 1$
$$x=\frac{mb(k+1) \pm b\sqrt{m^2(k+1)^2-4mk(m-1)}}{2(m-1)}$$
This expression is always defined except if the expression under the radical is negative.
Case 2a: $0<m<1$
Let's examine the radical expression, which if positive means $x$ is defined and the parabolas cross:
$$m(k+1)^2-4k(m-1)$$
The first term is positive, and the second term is positive unless $k<0$, meaning the roots are on opposite sides of $x=0$.
To avoid parabolas intersecting we must have $k<0$ and:
$$\frac{4k}{(k+1)^2} > \frac{m}{m-1}$$
The lower the value of $m$, the wider the range of $k$ that satisfy non-intersection. For example, if $m=0.5$, $0>k>-5$ satisfy, but if $m=0.9$, $k<-2$ are prohibited.
This case corresponds to a parabola "wider" than $F(x)$. Intuitively, we imagine if a wide parabola has a vertex mostly below and not too far to the side of $F(x)$ it might not cross, and the wider the parabola, the more leeway we have. That's exactly what the formulas give too.
Case 2b: $m>1$
$$m(k+1)^2-4k(m-1)$$
Again, in the radical expression, the first term is positive, and the second term is only negative when $k>0$, meaning that both roots are on the same side of $x=0$.
To avoid intersection this requires $k>0$ and:
$$\frac{4k}{(k+1)^2} > \frac{m}{m-1}$$
Inverting the fractions yields $\frac{(k+1)^2}{4k} < \frac{m-1}{m}$, the RHS is always $<1$, and the LHS is always $\geq 1$, so this case can never avoid intersection.
In this way, we see that the only ways for parabolas to avoid intersection are to be identical shapes with one vertically displaced downwards, or for the lower one to be "wider" than the first, and not too far to the side of the axis of the first, with more leeway the wider it is.
No, it does not. A simple counter-example is when $f(x)=2x^2-1$ and $g(x)=x^2.$ Although the minimum of $f(x)$ is $-1$ and the minimum of $g(x)$ is $0,$ $g(x)>f(x)$ does not hold for $x=2.$