Therefore
$$
a b x+a b x^{\prime}<a x+b x^{\prime}
$$
or
$$
a b<a x+b x^{\prime}
$$
On the other hand,
$$
\begin{aligned}
&(a<a+b)<[a x<(a+b) x] \
&(b<a+b)<\left[b x^{\prime}<(a+b) x^{\prime}\right]
\end{aligned}
$$
Therefore
$$
a x+b x^{\prime}<(a+b)\left(x+x^{\prime}\right)
$$
or
$$
a x+b x^{\prime}<a+b
$$
To sum up,
$$
a b<a x+b x^{\prime}<a+b .
$$
Q. E. D.
Remark 1. This double inclusion may be expressed in the following form:
$$
f(b)<f(x)<f(a) .
$$
For
$$
\begin{gathered}
f(a)=a a+b a^{\prime}=a+b \
f(b)=a b+b b^{\prime}=a b
\end{gathered}
$$
MATH3066 COURSE NOTES ：
The same conclusion may be reached by observing that abis the inferior limit of the function $a x+b x^{\prime}$, and that consequently the function can not vanish unless this limit is 0 .
$$
\left(a b<a x+b x^{\prime}\right)\left(a x+b x^{\prime}=0\right)<(a b=0) .
$$
We can express the resultant of elimination in other equivalent forms; for instance, if we write the equation in the form
$$
\left(a+x^{f}\right)(b+x)=0
$$
we observe that the resultant
$$
a b=0
$$
is obtained simply by dropping the unknown quantity (by suppressing the terms $x$ and $\left.x^{\prime}\right)$. A gain the equation may be written:
$$
a^{\prime} x+b^{\prime} x^{\prime}=1
$$