> [!NOTE] > Let $n,m>1$ and $x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_m$ be positive integers. Assume that $x_1+x_2+\cdots+x_n=y_1+y_2+\cdots+y_m<m n$. Prove that in the equality$x_1+x_2+\cdots+x_n=y_1+y_2+\cdots+y_m $ > one can suppress some (but not all) terms in such a way that the equality is still satisfied. ###### Proof Let $m=n=2.$ Then the possibilities are $1+1=1+1$ or $1+2=1+2$ and in either case we can suppress some (but not all terms) in a such a way that the equality is satisfied. Proceed by induction on $m+k.$ Suppose true for $m+n=k.$ Now suppose $m+n=k+1.$ WLOG suppose $x_{1}=\max_{i}x_{i}$ and $y_{1}=\max_{i}y_{i}$ and $x_{1} \geq y_{1}.$ If $m=2,$ then $2n>y_{1}+y_{2} = x_{1} +x_{2} +\dots+x_{n}\geq x_{1} +n-1 \geq y_{1} + n-1 $ Thus $y_{1} \geq y_{2}\geq n-1$ and $y_{1}+n-1 < 2n$ gives $y_{1}<n+1$ so: $y_{1}=x_{1}=n$ or $n-1,$ $y_{2}=n-1,$ $x_{2}=x_{3}=\dots=x_{n}=1,$ and we are done. If $m>2,$ then we can rewrite the original equality as $(x_{1}-y_{1})+x_{2}+\dots+x_{n}=y_{2}+\dots+y_{m}$Note that if $x_{1}-y_{1}=0$ then $x_{1}$ and $y_1$ are the numbers to be suppressed. We can apply the inductive hypothesis if $y_{1} \geq n,$ in which case $y_{2}+\dots+y_{m}< mn-y_{1} \leq (m-1)n.$ In this case, one can suppress the terms provided by the inductive hypothesis. Otherwise if $y_{1} <n,$ then $y_{2}+y_{3}+\dots+y_{m}\leq (m-1)y_{1}<(m-1)n$ and again the inductive hypothesis can be applied.