Proving a simple math inequality

Question

Let's say I have the following relations (all real numbers):

$$a_2 \leq a_1$$ $$x_1 \leq v_1$$ $$x_1 + x_2 \leq v_1 + v_2$$

How to show: $$a_1x_1+a_2x_2 \leq a_1v_1 + a_2v_2$$

This can be shown intuitionally but I am unable to show this mathematically. What can be done to show this holds? NOTE: Please take care of negative values too.

Answer 1

It's wrong.

Take $$(a_1,a_2,x_1,x_2,v_1,v_2)=(-2,-1,1,2,2,2).$$ We need to prove that: $$-2\cdot1+(-1)\cdot2\leq-2\cdot2+(-1)\cdot2$$ or $$-4\leq-6,$$ which is not true.

Your second problem is still wrong. Take: $$(a_2,a_1,x_1,x_2,v_1,v_2)=(-2,-1,1,2,2,2).$$

Answer 2

Let $\mathbf{a}=(a_1,a_2)^T$, $\mathbf{x}=(x_1,x_2)^T$, $\mathbf{v}=(v_1,v_2)^T$ and $\mathbf{p}=(1,0)^T$, $\mathbf{q}=(1,1)^T$.

We have $$\mathbf{p} \cdot \mathbf{x} \le \mathbf{p} \cdot \mathbf{v}$$ and $$\mathbf{q} \cdot \mathbf{x} \le \mathbf{q} \cdot \mathbf{v}$$

we can find $\alpha$ and $\beta$ so that $\mathbf{a}=\alpha\mathbf{p}+\beta\mathbf{q},$ namely $\alpha=a_1-a_2 \ge 0$ and $\beta=a_2$.

$$\alpha \ge 0 \Rightarrow$$ $$\alpha \,\mathbf{p}\cdot \mathbf{x} \le \alpha \,\mathbf{p}\cdot \mathbf{v} \phantom{12345}(*)$$

Now assume $\beta=a_2\ge 0:$

$$\beta \,\mathbf{q}\cdot \mathbf{x} \le \beta \,\mathbf{q}\cdot \mathbf{v} \phantom{12345}(**)$$

so, adding $(*)$ and $(**)$ $$( \alpha \mathbf{p} +\beta \mathbf{q} ) \cdot \mathbf{x} \le ( \alpha \mathbf{p}+\beta \mathbf{q} ) \cdot \mathbf{v}$$ or $$\mathbf{a} \cdot \mathbf{x} \le \mathbf{a} \cdot \mathbf{v}$$ which is what we wanted to show.

Answer 3

$$x_1 \leq v_1 \implies a_1x_1 \leq a_1v_1$$ $$a_2 \leq a_1 \implies a_2 x_2 \leq a_2v_2$$

Add these inequalities together.