Prove that $\sum_{i=1}^{n}a_i^2\geq\frac{A^2}{n}$, where $a_1,...,a_n\in\mathbb Z^+,\sum_{i=1}^{n}a_i=A$. Suppose we have $n$ positive integers $a_1,...,a_n$ and $\sum_{i=1}^{n}a_i=A$. If we further assume $a_i&lt;n$ for every $i$, then is the inequality $\sum_{i=1}^{n}a_i^2\geq\frac{A^2}{n}$ true? If yes, how can we prove it?
Just simply use the QM-AM inequality. It makes perfect sense as each $a_{i}$ $\in$ $\mathbb{N}$. Then by QM-AM inequality, we have that
\begin{align*}
&amp;\sqrt{\frac{\sum_{i= 1}^{n} a_{i}^{2}}{n}} ~\geqslant~\frac{\sum_{i = 1}^{n}a_{i}}{n}\\
\Leftrightarrow~ &amp; \frac{\sum_{i= 1}^{n} a_{i}^{2}}{n} ~\geqslant~ \frac{A^2}{n^2}\\
\Leftrightarrow~ &amp; {\sum_{i= 1}^{n} a_{i}^{2}} ~\geqslant~ \frac{A^2}{n} 
\end{align*}
By Jensen's inequality, since $\frac{d^2}{dx^2}x^2&gt;0$,
$$\frac{\displaystyle\sum_{i=1}^na_i^2}n\ge \left(\frac{\displaystyle\sum_{i=1}^na_i}n\right)^2$$
, where $a_i\in\mathbb R,n\in\mathbb Z^+$.