how does the rounded step came in the following image??? please help... i am new to this forum.. please forgive if anything is wrong..
sorry sir.. now i have attached the image...
http://images.elektroda.net/88_1304862170.png
Interesting... I think I came up with a proof, but it is way too complicated.
We need to show that $\displaystyle \sum_{i=1}^k n_i^2\le n^2-(k-1)(2n-k)$. The left-hand side is $\displaystyle n^2-2\sum_{1\le i<j\le k}n_in_j$, so it is sufficient to show that $\displaystyle 2\sum_{1\le i<j\le k}n_in_j\ge (k-1)(2n-k)$.
The following statement can be proved by induction on k: For every sequence $\displaystyle n_1,\dots,n_k$ where each $\displaystyle n_i\ge 1$ (and, therefore, $\displaystyle k\le n$), we have $\displaystyle 2\sum_{1\le i<j\le k}n_in_j\ge (k-1)(2n-k)$ where $\displaystyle n=n_1+\dots+n_k$.
However, I am really wondering if there is an easier way.
If you need more details, feel free to ask.
Edit: If you don't mind, where does this proof come from?
Above, I indicated the steps to prove $\displaystyle \sum_{i=1}^k n_i^2\le n^2-(k-1)(2n-k)$. The key lemma is proved by induction. If you have questions about these particular steps, including the proof by induction, feel free to ask, but your questions have to be more specific.
The proof suggested above looks too complicated for the transition from one line to the next in the original post. On the other hand, some authors sometimes do take out a whole page of calculations and replace it with the words, "It is obvious that..." I would really like to know if this is the case here.