For a tournament T of order n, let $Δ=max \{od v:v∈V(T)\}$ And $δ=min\{od v:v∈V(T)\}$ Prove that if $Δ-δ< \frac n2$, then T is strong

Question

For a tournament $T$ of order $n$, let $Δ=max \{od v:v∈V(T)\}$ And $δ=min\{od v:v∈V(T)\}$ Prove that if $Δ-δ< \frac n2$, then $T$ is strong

Here is my final attemp. Prove this by contrapositive.

Assume that $T$ is not strong, by the result of some previous exercise, $T$ is not regular. So there exists a vertex $v$ such that $id(v) <od(v)$. Note that $T$ is a tournament which is an oriented completed graph so

$$deg(v)=id(v)+od(v)=n-1$$

Since $id(v)<od(v)$, $ \delta \leq od(v)< \frac {n-1}2$

Let $\pi: s_1,s_2,...,s_n$ be the score sequence of $T$, then

$$\Delta= s_n= \sum_{i=1}^n S_i -\sum_{n=1}^{n-1}S_i \geq \left( \begin{array}{ccc} n \\ 2 \\ \end{array} \right) -\left( \begin{array}{ccc} n-1 \\ 2 \\ \end{array} \right)=n-1$$

So

$$\Delta -\delta >(n-1) -\frac {n-1}2= \frac {n-1}2 \geq \frac n2$$

By contrapositive, if $Δ-δ< \frac n2$, then $T$ is strong

Is my reasoning acceptable?

Answer 1

\frac{n-1}{2} is not bigger than \frac{n}{2}