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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.

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  • $\begingroup$ The last result is still not correct $\endgroup$ Commented Oct 14, 2020 at 4:17
  • $\begingroup$ 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. $\endgroup$
    – mjw
    Commented Oct 14, 2020 at 4:22
  • $\begingroup$ See my answer. You can have a weaker constraints than that. $\endgroup$ Commented Oct 14, 2020 at 4:26
  • $\begingroup$ On the LHS you need $a_2x_1 \dots$ $\endgroup$ Commented Oct 14, 2020 at 4:29
  • $\begingroup$ I'm afraid you're right. Sorry. I'll fix it. $\endgroup$ Commented Oct 14, 2020 at 4:32

3 Answers 3

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

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

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  • $\begingroup$ Please check the edited question. I am sorry I posted the wrong inequality. $\endgroup$
    – DuttaA
    Commented Oct 14, 2020 at 4:10
  • $\begingroup$ Okay.${}{}{}{}{}{}$ $\endgroup$
    – mjw
    Commented Oct 14, 2020 at 4:20
  • $\begingroup$ p.x is not greater than p.v .... $\endgroup$
    – DuttaA
    Commented Oct 14, 2020 at 5:43
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$$x_1 \leq v_1 \implies a_1x_1 \leq a_1v_1$$ $$a_2 \leq a_1 \implies a_2 x_2 \leq a_2v_2$$

Add these inequalities together.

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  • $\begingroup$ What happens if $a_1<0$? $\endgroup$
    – mjw
    Commented Oct 14, 2020 at 4:31
  • $\begingroup$ I assumed the scalars were positive. $\endgroup$ Commented Oct 14, 2020 at 4:34

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