Prove that chromatic number $\geq \frac{V^2}{V^2-2E}$ Let us have a simple graph G. (That means no loops and no multiple edges)
Prove that $\chi \geq  \frac{V^2}{V^2-2E}$
I found the task in the book by John Bondy "Graph Theory with Applications" (Task 8.1.1)
The preliminary theorems didn't help me, and trying to transform the equation didn't help either.
I know that $2E \geq \chi \cdot (\chi -1)$. And that $V \geq \chi$
My equation can be transformed to $V^2 \geq 2E \cdot \frac{\chi}{\chi -1}$
Am I close?
Searching I have discovered that my question is a duplicate.
The answer down here is partial, but more beautiful than the answer to the question, which is duplicated, though.
 A: Case 1. $V \geq 2E$.
Observation. If $G$ is simple, $\binom{V}{2} \geq 2E$. i.e., $V^2-2E \geq V$.
For $\chi=1$, $E=0$, and $G$ is an isolated point.
Now, WMA $\chi\geq 2$. (So, $E>0$)
Then, by the above observations, $1 \leq (\chi-1)\cdot\frac{V}{2E} \leq (\chi-1)\cdot\frac{V^2-2E}{2E}$.
Thus, $(\chi-1)\cdot(V^2-2E)\geq 2E$
$\Leftrightarrow$ $(\chi-1)V^2 \geq 2E\cdot\chi$
$\Leftrightarrow$ $V^2\geq 2E\cdot\frac{\chi}{\chi-1}$.
Case 2. $V<2E$.
Unfortunately, I have no idea for this case.
----Edit.----
Rather than above incomplete proof, I suggest other proof.
Now, add one more condition that $G$ is connected.
Case 1. $\chi\leq 2$. $G$ is bipartite. Most possible number of edges is $E_\max 
 = \lfloor \frac{V}{2} \rfloor \lceil \frac{V}{2} \rceil$. Then,
$\frac{V^2}{V^2-2E} \leq \frac{V^2}{V^2-2E_\max} \leq 2$.
Case 2. $\chi\geq 3$. Since $G$ is connected and $\chi \geq 3$, $V\geq E$. Thus, we can observe that
$1 \leq (\chi-1)\cdot\frac{V}{2E} \leq (\chi-1)\cdot\frac{V^2-2E}{2E}$.
Now, follow my original incomplete proof.
I think that we can do for $G$: not connected.
