Here is a problem I found recently:
If $a\ge{b}$,$x\ge{y}$, prove that $ax+by\ge{ay+bx}$ for all real numbers $a,b,x,y$.
I found $2$ solutions to the problem, which possibly contradict each other.
Solution 1
We have $(a-b)\ge{0}$ and $(x-y)\ge{0}$. Multiplying them with each other, we have $$(a-b)(x-y)\ge{0}$$ $$\Rightarrow{ax+by}\ge{ay+bx}$$ as desired.
Solution 2
There are several cases when for positive and negative $a,b,x,y$. We consider the case when $a,b$ are positive and $x,y$ are negative. Then we have $$ax\le{bx}\cdots(1)$$ $$ay\le{by}\cdots(2)$$ From $(1)-(2)$, we have $$ax-ay\le{bx-by}$$ $$\Rightarrow{ax+by}\le{ay+bx}$$
The above 2 solutions contradicts each other. Is there something wrong in my solutions?
Note: The question seems lame comparing it with other questions in MSE. But I'm a grade-4 student and I stucked at this question. So, please help me and ignore my mistakes.
First method is flawless.
But the second you correctly that if $b \ge a> 0$ and $y \le x < 0$ then $ay \le by$ and $ax \le bx$
But when we says $1- 2$ implies..... you have no rule that if $M < N$ and $J < K$ that $M -J < N-K$. As $J < K\implies -J > -K$ we have no way of comparing $M +(-J)$ to $N + (-K)$ when $M< N$ but $-J > -K$.
Indeed we actually have $ax - ay = a(x-y)$ and as $x-y > 0$ we have $a(x-y) \ge b(x-y) = bx -by$.