How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions?
I obtained the definition of the principle of variable proportion from this site
It states:

... means that upto the use of a certain amount of variable factor, marginal product of the factor may increase and after a certain stage it starts diminishing.

The following is a technical derivation. See the picture at the end for an illustration of the various results.
Denote by $X$ the variable input and by  $Q = {\cal Q}(X)$ the output produced by the amount $X$. The marginal product can be expressed using the first derivative $MP(X) = {\cal Q}'(X)$. The increase or decrease of the marginal product then says something about the second order derivative $MP'(X) = {\cal Q}''(X)$.
The principle of variable proportions then states that $MP'(X)$ is first positive and then negative.




period
change
sign of $MP(X) = {\cal Q}'(X)$
sign of $MP'(X) = {\cal Q}''(X)$




1
marginal product increases
$&gt;0$
$&gt; 0$


2
marginal product decreases
$&gt; 0$
$&lt; 0$




Given that ${\cal Q}'(X) &gt; 0$ we can invert ${\cal Q}(X)$ to obtain  $X = {\cal X}(Q)$ which gives the amount of variable input that produces $Q$ Then as the following identity holds for all $X$:
$$
{\cal X}(Q(X)) = X,
$$
We can take the derivative on both sides and obtain:
$$
{\cal X}'({\cal Q}(X))\,\, {\cal Q}'(X) = 1 \iff {\cal X}'(Q) = \frac{1}{{\cal Q}'(X)}.
$$
and taking derivatives once more gives:
$$
{\cal X}''({\cal Q}(X)) \,\,({\cal Q}'(X))^2 + {\cal X}'({\cal Q}(X))\,\,{\cal Q}''(X) = 0 \iff {\cal X}''(Q) = -\frac{{\cal X}'({\cal Q}(X))\,\, {\cal Q}''(X)}{({\cal Q}'(X))^2}.
$$
This allows us to complete the table:




period
change in $Q$
sign of ${\cal Q}'(X)$
sign of ${\cal Q}''(X)$
sign of ${\cal X}'(Q)$
sign of ${\cal X}''(Q)$




1
marginal product increases
$&gt;0$
$&gt; 0$
$&gt; 0$
$&lt;0$


2
marginal product decreases
$&gt; 0$
$&lt; 0$
$&gt; 0$
$&gt;0$




Now, let's have a look at the variable cost. If $w$ is the unit price of an input $X$,  then total variable costs is:
$$
VC(Q) = w {\cal X}(Q).
$$
Marginal costs is the derivative of the variable costs:
$$
MC(Q) = VC'(Q) = w {\cal X}'(Q)
$$
Th Marginal costs are first decreasing and then increasing.

proof: follows from the previous table as ${\cal X}''(Q)$ is first negative and then positive.

Average variable cost is variable costs divided by output:
$$
AVC(Q) = \frac{w {\cal X}(Q)}{Q}.
$$
Now let's have a look at how the  $AVC(Q)$ varies with $Q$ (which can be expressed using the derivative $AVC'(Q)$). We have:
$$
AVC'(Q) = \frac{(w {\cal X}'(Q))Q  - (w{\cal X}(Q))}{Q
^2} = \frac{MC(X) - AVC(Q)}{Q},\\
$$
Th $AVC(Q)$ is decreasing if  $MC(Q) &lt; AVC(Q)$ and increasing if $MC(Q) &gt; AVC(Q)$. If $AVC(Q)$ is at a minimum (or maximum) then $MC(Q) = AVC(Q)$.

proof: follows from the equation above as the sign of $AVC'$ depends on the sign of $MC - AVC$.

The following shows that for $Q = 0$, average and marginal costs are identical.
Th If $MC(0)$ exists, then $MC(0) = AVC(0)$.

proof: We have that $AVC(Q) = w {\cal X}(Q)/Q$. Using l'Hospital, when taking the limit of $Q \to 0$, gives  $AVC(0) = w {\cal X}'(0) = MC(0)$.

The following shows that initiall, for low values of $Q$, the $AVC$ will be decreasing.
Th for $Q$ small enough, $MC &lt; AVC$, and therefore $AVC$ is decreasing.

proof, Assume not, then $MC &gt; AVC$ in some neighbourhood $]0, Q[$ which means that $AVC$ is increasing. Also, we know that $MC$ is initially decreasing. As $MC(0) = AVC(0)$, this means that $AVC &gt; MC$ in $]0,Q[$, a contradiction.

The following shows that the $MC$-curves cuts the $AVC$-curve from below.
Th let $Q^\ast$ be the first point where $MC(Q^\ast) = AVC(Q^\ast)$, then $MC'(Q^\ast) &gt; 0$ , so the $MC$-curve cuts the $AVC$-curve from below.

proof: From the previous result, we know that slightly before $Q^\ast$, $MC &lt; AVC$  and at $Q^\ast$, $MC = AVQ$ so $(MC - AVQ)' &gt; 0$ at $Q^\ast$.  Working out this derivative, shows that $MC' - \frac{MC - AVC}{Q} &gt; 0$. As $MC = AVC$ at $Q^\ast$, we have $MC'(Q^\ast) &gt; 0$.

The following shows that the $MC$-curve cuts the $AVC$-curve at most once.
Th let $Q^\ast$ be the first point where $MC(Q^\ast) = AVC(Q^\ast)$ and let $Q &gt; Q^\ast$  then $MC(Q) &gt; AVC(Q)$ and therefore $AVC$ is increasing. As such, there is at most one point where the $MC$ is equal to the $AVC$.

proof: As at $Q^\ast$ , $MC$ is in its increasing part, we have (from the table above) that $MC(Q) &gt; 0$ for all $Q \ge Q^\ast$,  Also, as  $AVC(Q^\ast) = MC(Q^\ast)$ and $MC'(Q^\ast) &gt; 0$ and $AVC'(Q^\ast) = 0$, we have for $Q &gt; Q^\ast$ close enough to $Q^\ast$ that $MC(Q) &gt; AVC(Q)$. So $AVC(Q)$ is increasing for all $Q&gt; Q^\ast$ close enought to $Q$.
Next, for such $Q$,  $(MC(Q) - AVC(Q))' = MC'(Q) + (MC(Q) - AVC(Q))/Q &gt; 0$, so the distance between $MC(Q)$ and $AVC(Q)$ is increasing. this means that the distance between $MC(Q)$ and $AVC(Q)$ will grow bigger and bigger. As such, the two curves will never again be equal so $MC(Q) &gt; AVC(Q)$ for all $Q &gt; Q^\ast$.

The final situation can be seen in the picture below. The $MC$ curve is first decreasing and then increasing. At $Q = 0$ the two are the same. First the $MC$ curve obtains its minimum. Then in its increasing part, it cuts the $AVC$ curve at the point where the $AVC$ curve is at its minimum. After this point the distance between the $MC$ and $AVC$ curve grows bigger and bigger, so the two curves never intersect again, and both courves keep increasing.