Easiest way to solve this inequality? $m^2-mn+n^2+m-n>0$ for $m,n \in \mathit Z^+$

Question

The problem in question is to prove:

$m^2-mn+n^2+m-n>0$ for $m,n \in \mathit Z^+$.

What is the easiest/most consistent way to deal with such problems? Is there any identity that can be applied when we only care about the positive integers?

Thanks.

Answer 1

$$ m^2 -mn +n^2 + m -n = (m-n)^2 + mn + m -n = (m -n)^2 + n(m-1) + m $$ Since $m,n \in Z^+$, the first and second terms are $\ge 0$, the third is $\gt 0$ so the sum is $\gt 0$.

This is certainly a special case treatment; but even though there are some common idioms here, I don't think you can come up with a general way to treat such problems.

Answer 2

The hint:

$$m^2-mn+n^2+m-n=\frac{1}{2}((m+1)^2+(n-1)^2+(m-n)^2)-1.$$ Now, prove that: $$(m+1)^2+(n-1)^2+(m-n)^2\geq4.$$

Answer 3

Playing around, as usual.

$\begin{array}\\ m^2-mn+n^2+m-n &=\frac12(2m^2-2mn+2n^2)+m-n \qquad\text{I've done these steps before}\\ &=\frac12(m^2-2mn+n^2+m^2+n^2+2m-2n)\\ &=\frac12((m-n)^2+m^2+2m+n^2-2n)\\ &=\frac12((m-n)^2+m^2+2m+1+n^2-2n+1-2)\\ &=\frac12((m-n)^2+(m+1)^2+(n-1)^2-2)\\ &\ge\frac12((2)^2-2) \qquad\text{use } m \ge 1\\ &= 1\\ \end{array} $