Theorem
Let $$f_1(x)$$, $$f_2(x)$$, and $$g(x)$$ be functions of $$x$$ over the real interval $$X\equiv[A,B]$$. Also let $$x_1,x_2\in[A,B]$$. If:
• For all $$x\in[0,B]$$, $$f_2(x) \geq f_1(x)$$,
• $$f_1(x)$$ is increasing, $$g(x)$$ is continuous,
• $$f_2(A) \geq f_1(A) > g(A)$$,
• For each $$i$$, and $$\forall x \in X, f_i(x)=g(x)\Leftrightarrow x=x_i$$,
Then $$x_2\geq x_1$$
###### Proof
If $$x < x_1$$, then $$f_1(x)>g(x)$$.
Let $$\dot{x} \equiv \inf \left\{ x\in X \mid f_1(x) \leq g(x) \right\}$$.
If $$x < \dot{x}$$, then $$f_1(x)>g(x)$$ .
It must be that $$\dot{x}=x_1$$.
Note that $$\dot{x} > A$$
By the continuity of $$g(x)$$, there is an $$\varepsilon > 0$$ such that $$g(x)< f_1(A)$$ for all $$x \leq A+\varepsilon$$.
But $$f_1$$ is increasing, so for all $$x\leq A+\varepsilon$$, $$f_1(x)\geq f_1(A)$$.
Thus for all $$x\leq A+\varepsilon$$, $$f_1(x) > g(x)$$. So $$\dot{x} \geq A+\varepsilon$$.
Note that if $$x < \dot{x}$$, then $$f_1(\dot{x})\geq f_1(x) > g(x)$$.
The set of $$x<\dot{x}$$ is nonempty, so by the continuity of $$g(x)$$, $$f_1(\dot{x})\geq g(\dot{x})$$.
If $$f_1(\dot{x})> g(\dot{x})$$, then a contradiction arises:
By the continuity of $$g(x)$$, there is an $$\varepsilon > 0$$ such that $$f_1(\dot{x}) > g(x)$$ for all $$x\in B_\varepsilon [\dot{x}]$$.
But $$f_1$$ is increasing, so for all $$x\geq\dot{x}$$, $$f_1(x)\geq f_1(\dot{x})$$.
Thus for all $$x\leq \dot{x}+\varepsilon$$, $$f_1(x)\geq f_1(\dot{x}) > g(x)$$.
And so $$\dot{x}$$ is not the infimum of $$\left\{ x\in X \mid f_1(x) \leq g(x) \right\}$$. ↯.
Therefore $$f_1(\dot{x})=g(\dot{x})$$, which implies that $$\dot{x}=x_1$$.
So if $$x < x_1$$ then $$f_2(x)\geq f_1(x) > g(x)$$.
And $$f_2(x_2)=g(x_2)$$, so by contrapositive, $$x_2 \geq x_1$$.
###### But also, I mean, just look at it.
 If you shine a beam ($$g$$) up at two continuous barriers, and one barrier ($$f_1$$) is always on top of the other ($$f_2$$), then the beam will strike the barrier on the bottom first. To strike the top barrier first, the beam has to make it through a gap in the bottom barrier without striking it. This kind of gap doesn't work because $$f_1$$ is increasing. And this kind of gap doesn't work because to fly through it, the beam would have to travel to the left, which it can't do because it's a continuous function of $$x$$. Even if we wiggle the beam a bit, it can't go left. So even if we allow for discontinuities, the beam can't get through.