Elementary Inequality regarding Sum of Squares

begin{align} &mbox{If}quad A = sum_{i = 1}^{n}a_{i},x_{i}^{2} quadmbox{and}quad B = sum_{j = 1}^{n}b_{j},y_{j}^{2} \[2mm] end{align}

with integer $x_{i}, y_{j}, a_{i}, b_{j} geq 1 , quad x_{i} neq y_{j} $, and $A > B$, we are to find conditions for which $A-B > 1$.

As OP already posted, the given condition $A > B$, together with the integer condition of the variables, implies that the difference $A - B$ cannot become arbitrarily small. So we start to investigate the smallest such difference under the given conditions.

This can be used to construct first of all an example where the assertion does not hold true. Let $A = 1cdot 3^2 = 9$ and $B = 2cdot 2^2 = 8$, then $A>B$ and $A-B=1$ so the required $A-B>1$ fails. There are many more such examples possible for other integer numbers $x_{i}, y_{j}, a_{i}, b_{j} geq 1$. So, the only statement we can make in general is that, due to the integer condition and $A>B$, the minimum possible difference is $A-B=1$.

From this it follows that it's difficult resp. unlikely to give general explicit conditions for making $A-B> 1$. Such conditions can be implicitely stated: Choose integer numbers $x_{i}, y_{j}, a_{i}, b_{j} geq 1$ such that $A-Bge 2$. (Then $A-B > 1$ is obviously obtained.)