Show that the size of the Turan graph $T_r(n)$ is at least $(1 - \frac{1}{r}) \binom{n}{2}$. A Turan graph $T_r(n)$ is defined as the complete $r$-partite graph of order $n$ such that the number of vertices in each of the $r$ classes is either $\lfloor \frac{n}{r}\rfloor$ or $\lceil \frac{n}{r} \rceil$. For fixed $n$ and $r$, $T_r(n)$ is unique up-to isomorphism. The size of $T_r(n)$ can be simply counted as: $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$.
Here is what I have: assume $n = kr + s,\ 0 \leq s \leq r-1$. Note that at least one class must have exactly $\lfloor \frac{n}{r} \rfloor$ vertices. Then, $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$
$\geq$ $\binom{n}{2} -(r-1) \binom{\lceil \frac{n}{r} \rceil}{2} - \binom{\lfloor \frac{n}{r} \rfloor}{2}$
$=\binom{n}{2} - (r-1) \binom{k+1}{2} - \binom{k}{2}$
$= \frac{n(n-1)}{2} - \frac{(r-1)k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)}{2} - \frac{rk(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$\geq \frac{n(n-1)}{2} - \frac{n(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + k(k+1) - k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + 2k}{2}$
$= \binom{n}{2}(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$.
But clearly, $(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$ may be smaller than $1 - \frac{1}{r}$, as seen by taking $n=31$ and $r=5$.

The size $S$ of the Turán graph is $\frac 12\left(n^2-\frac {(n^2-s^2)}{r}-s\right)$, see my answer. It follows $$S- \left(1 - \frac{1}{r}\right) \binom{n}{2}=\frac{(n-s)(r-1)+s(s-1)}{2r}\ge 0.$$
