In an equilateral triangle, prove that $|BQ| + |PQ| + |CP| > 2l$ I am trying to solve the following problem:

Let $ABC$ be an equilateral triangle with side $l$.
If $P$ and $Q$ are points respectively in sides $AB$ and $AC$, different from the triangle vertices, prove that $$|BQ| + |PQ| + |CP| > 2l$$

I can see that, as point $P$ tends to $A$, $|CP|+|PQ|$ tends to $|AC|+|AQ|$. If I could prove this, the problem would be solved (the rest follows from the triangle inequality).
However I have no clue on how to do this. I tried to play with triangle inequality and relations between sides and angles but nothing worked.
How can I proceed?
 A: Let $AQ=x$, $AP=y$ and $l=1$.
Thus, $$PQ=\sqrt{x^2-xy+y^2},$$
$$PC=\sqrt{y^2-y+1}$$ and
$$BQ=\sqrt{x^2-x+1}$$ and we need to prove that
$$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$
Now, by Minkowwski
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}\geq$$
$$\geq\sqrt{\left(x-\frac{1}{2}+y-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=\sqrt{(x+y-1)^2+3}$$ and
$$\sqrt{x^2-xy+y^2}\geq\frac{x+y}{2}.$$
Let $x+y=2a$.
Hence, $a\leq1$ and we need to prove that
$$a+\sqrt{(2a-1)^2+3}\geq2$$ or
$$\sqrt{4a^2-4a+4}\geq2-a$$ or
$$4a^2-4a+4\geq a^2-4a+4,$$
which is obvious.
Done!
By my solution easy to make a geometric proof.
A: Upon reflection ...

$$2s = |\overline{AB^\prime}| \leq |\overline{AP}|+|\overline{PQ^\prime}|+|\overline{Q^\prime B^\prime}| = p+q+r$$
A: This is not an answer, rather it is a suggested approach to the problem. Here is a Geogebra (link)diagram:
If you go to the Geogebra link you can move points $P$ and $Q$.
The problem is reduced to proving that $|DI|>|IQ|$.

Note from the construction that $|PQ|=|DQ|$ and $|CP|=|BG|=|BF|$ and that $2l=|IH|$.
Therefore
$$ |PQ|+|BQ|+|CP|=|DQ|+|QB|+|BF|=|DF| $$
So to establish the result it must be shown that
$$ |DI|>|FH|=|IQ| $$
A: Refer to the graph (using smartphone, will draw at PC later) below:

