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If $f,g\colon X\to \mathbb{R}$ are continuous at $a \in X$ and $f(a)<g(a)$, then there is a $\delta>0$ such that $f(x)<g(x)$ for all $x\in X$ with $\left|x-a\right|<\delta$.

If $f$ and $g$ are continuous at $a\in X$, then for all $\varepsilon>0$ there is a $\delta_f>0$ and a $\delta_g>0$ such that for all $x\in X$ we have

$$\left|x-a\right|<\delta_f\implies \left|f(x)-f(a)\right|<\varepsilon$$ as well as $$\left|x-a\right|<\delta_g\implies \left|g(x)-g(a)\right|<\varepsilon$$

Choose $\varepsilon=g(a)-f(a)$, there there'll be $\delta_f,\delta_g>0$ and let's define $\delta=\min{(\delta_f,\delta_g)}$.

Then we'll have for all $x\in X$ that $$\left|x-a\right|<\delta\implies \left|g(x)-g(a)\right|<\varepsilon \ \text{and}\ \left|f(x)-f(a)\right|<\varepsilon$$

Since $\varepsilon=g(a)-f(a)$, we have

$$\left|g(x)-g(a)\right| <g(a)-f(a) \implies f(a)<g(x)$$

and

$$\left|f(x)-f(a)\right| <g(a)-f(a) \implies f(x)<g(a)$$

But from now I don't really know how to arrive at $f(x)<g(x)$, as $f(a)<g(x)$ and $f(x)<g(a)$ don't really imply much.

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    $\begingroup$ Take $\varepsilon=\frac 12(g(a)-f(a))$ instead. $\endgroup$ – August Liu Nov 17 '20 at 15:18
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You can simplify a lot the proof by denoting $h(x) = g(x) - f(x)$. You have $h(a) >0$ and have to prove that $h(x) >0$ in a neighborhood of $a$.

By continuity of $h$ at $a$, it exists $\delta > 0$ such that $\vert h(x) -h(a)\vert < \frac{ h(a) }{2}$ for $\vert x-a \vert < \delta$.

As $\vert h(x) -h(a)\vert < \frac{ h(a) }{2}$ implies $h(x) > \frac{h(a)}{2} >0$, you're done.

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