# Proving a simple math inequality

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.

• The last result is still not correct Commented Oct 14, 2020 at 4:17
• Seems $x_1,x_2,v_1,v_2$ are free to be positive or negative but we require another condition such as $0\le a_2\le a_1$ for this to be true, in general.
– mjw
Commented Oct 14, 2020 at 4:22
• See my answer. You can have a weaker constraints than that. Commented Oct 14, 2020 at 4:26
• On the LHS you need $a_2x_1 \dots$ Commented Oct 14, 2020 at 4:29
• I'm afraid you're right. Sorry. I'll fix it. Commented Oct 14, 2020 at 4:32

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: $$(a_2,a_1,x_1,x_2,v_1,v_2)=(-2,-1,1,2,2,2).$$

• Sorry I posted the wrong inequality. The first inequality will be reverse. Thnak you very much for pinting this out. Commented Oct 14, 2020 at 4:06
• @DuttaA You new problem is also wrong. Commented Oct 14, 2020 at 4:13
• If we assume $a_1\ge a_2 \ge 0$, then it should work out.
– mjw
Commented Oct 14, 2020 at 4:19
• @MichaelRozenberg, Okay. I like your approach, though. A counter-example for each attempted problem formulation $\cdots$
– mjw
Commented Oct 14, 2020 at 4:27
• @DuttaA, We need $a_2 \ge 0.$
– mjw
Commented Oct 14, 2020 at 13:33

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.

• Please check the edited question. I am sorry I posted the wrong inequality. Commented Oct 14, 2020 at 4:10
• Okay.${}{}{}{}{}{}$
– mjw
Commented Oct 14, 2020 at 4:20
• p.x is not greater than p.v .... Commented Oct 14, 2020 at 5:43

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

• What happens if $a_1<0$?