If a unitsquare is partitioned into 101 triangles, is the area of one at least 1%?
As suggested in a comment above, I had asked a version of this question years ago. It makes sense to look at upper and lower bounds for the quantity $f(n)-\frac1n$. It is easily seen that $$ f(n) - \frac1n \le \frac1{n^2-n} $$ and indeed has shown that $$ f(n) - \frac1n = O\big(\frac1{n^3}\big), $$ see Bernd Schulze's paper "On the area discrepancy of triangulations of squares and trapezoids", The Electronic Journal of Combinatorics 18(1) (2011), #P137, 16 pp.
Computational experiments from an unpublished Diploma thesis (Katja Mansow, Ungerade Dreieckszerlegungen, TU Berlin 2003, 49 pages, in German) suggest that the true order of this difference is, however, singly-exponentially small. From gap-theorems in semi-algebraic geometry there is a doubly-exponential lower bound (details to be worked out).
For all $n \ge 5$ (odd) we have $f(n) < \frac{1}{n-1}$. So the answer to the question in the title is "not necessarily". Naturally, this does not answer the more interesting question on what the optimal $f$ is.
A triangulation showing this is in the picture: take a triangle $A$ of area $\frac{1}{n}$ at the left corner and divide the rest of the square with triangles so that (for instance)
the sides of the triangles that are partly common with the right side of the square have equal length and
the triangles having part of one of their side common with $A$ have equal area.
Are you sure about $n = 5$?
Start with a unit square and remove from it a right triangle of side $\epsilon$ and height 1. You are left with a trapezoid. Divide the trapezoid height-wise into $k$ pieces each of the same area, and chop each sub-trapezoid diagonally into two triangles. Each of the triangles will have area $\approx \frac{1 - \epsilon/2}{2k}$ (but not exact). First optimise to get
$$ \frac{\epsilon}{2} = \frac{1 - \epsilon /2}{2k} \implies \epsilon = \frac{2}{2k+1} $$
It is easy to see that the subtrapezoid with the two horizontal lengths being $1 - \epsilon$ and $1 - \epsilon + h\epsilon$ where $h$ is chosen so that
$$ h \cdot \frac{2 - 2\epsilon + h\epsilon}{2} = \epsilon \tag{h}$$
will be the one divided "most unevenly" by the diagonal cut ($h$ is largest, and so the difference in area between the two triangles is largest). Let us compute the areas of the two sub-pieces. The larger of the two pieces will have area
$$ h \cdot \frac{1 - \epsilon + h\epsilon}{2} = \frac{\epsilon}{2} (1 + \frac{h^2}{2}) $$
by (h). The quadratic formula for (h) implies
$$ h = \frac{1}{\epsilon} \left( \sqrt{1 - 2\epsilon + 3\epsilon^2} - (1-\epsilon)\right) \implies h < \frac{\epsilon}{(1-\epsilon)} = \frac{2}{2k-1}$$
and hence
$$ 1 + \frac{h^2}{2} < \frac{(2k-1)^2 + 2}{(2k-1)^2} $$
For $k = 2$ ($n = 5$) this implies that the largest piece in this decomposition has area
$$ \frac{\epsilon}{2}(1 + \frac{h^2}2) < \frac{1}{5} \frac{11}{9} < \frac{11}{44} = \frac14 $$
So your estimate is not sharp.
Note also that asymptotically the upper bound to the correction term
$$ 1 + \frac{h^2}{2} < 1 + \frac{2}{(2k-1)^2} < 1 + \frac{1}{2k} $$
for all $k> 2$. In fact, the correction decays much faster than the desired correction $1/2k$. So as $n$ gets large the difference between $f(n)$ and $1/(n-1)$ get proportionally bigger. The above example also shows that asymptotically $f(n)$ should approach $1/n$, with a correction that is of order $O(1/n^3)$.