Let $x_1,x_2,x_3,...,x_{19}$ be positive integers satisfying $\sum\limits_{i=1}^{19}x_i=2020$ and $x_i \geq 2$ for $i=1,2,...,19$.
Find the smallest value of $$P=\sum\limits_{i=1}^{19} x_i^2.$$
I've thought it was an easy question at the first place by using a very common result $$P \geq \dfrac{\left(\sum\limits_{i=1}^{19} x_i\right)^2}{19}=\dfrac{2020^2}{19}.$$
However, the equality does not occur since $x_i$ are positive integers. I'm struggling with finding another strategy for this problem! Many thanks.
The minimum value will be if all $x_i$ are ,,as equal as they can be*'', so if all are around 106. Say $k$ of them are exactly 106 and the rest of them 107, so we have $$106k+107(19-k)=2020$$ Now calculate $k$ and you are done.
This * is because of this inequality $$(x+1)^2+(y-1)^2\geq x^2+y^2$$ if $x>y$.