Proof: By Euclid
(related to Proposition: 7.39: Least Number with Three Given Fractions)
 For let $D$, $E$, and $F$ be numbers having the same names as the parts $A$, $B$, and $C$ (respectively).
 And let the least number, $G$, measured by $D$, $E$, and $F$, have been taken [Prop. 7.36].
 Thus, $G$ has parts called the same as $D$, $E$, and $F$ [Prop. 7.37].
 And $A$, $B$, and $C$ are parts called the same as $D$, $E$, and $F$ (respectively).
 Thus, $G$ has the parts $A$, $B$, and $C$.
 So I say that ($G$) is also the least (number having the parts $A$, $B$, and $C$).
 For if not, there will be some number less than $G$ which will have the parts $A$, $B$, and $C$.
 Let it be $H$.
 Since $H$ has the parts $A$, $B$, and $C$, $H$ will thus be measured by numbers called the same as the parts $A$, $B$, and $C$ [Prop. 7.38].
 And $D$, $E$, and $F$ are numbers called the same as the parts $A$, $B$, and $C$ (respectively).
 Thus, $H$ is measured by $D$, $E$, and $F$.
 And ($H$) is less than $G$.
 The very thing is impossible.
 Thus, there cannot be some number less than $G$ which will have the parts $A$, $B$, and $C$.
 (Which is) the very thing it was required to show.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"