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Suppose I have two inequalities: $$k_1 \leq a$$ $$k_2 \leq b$$ where $k_1,k_2,a,b$ are all positive numbers I know that that summation of them can be written as: $$k_1+k_2 \leq a+b$$ But I want to find $$k_1-k_2 \leq ?$$ or $$? \leq k_1-k_2$$ Is there any way to solve this problem? |
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First thing you can do is invoke the triangle inequality, i.e. $|x + y| \leq |x| + |y|$. Note that this equality works for negative $x$, $y$ as well. This gives you $$ \begin{eqnarray} |k_1 - k_2| &\leq& |k_1| + |k_2| &\leq |a| + |b| \text{ and since }k_1-k_2 \leq |k_1-k_2| \\ k_1 - k_2 &\leq& |k_1| + |k_2| &\leq |a| + |b| \end{eqnarray} $$ For arbitrary $k_1 \leq a$, $k_2 \leq b$, you can't do better than that, but since you stated that $k_1,k_2 \geq 0$, you can. $k_1,k_2$ being positive means that $$ \begin{eqnarray} -b &\leq& -k_2 &\leq& k_1 - k_2 &\leq& k_1 &\leq& a\\ -a &\leq& -k_1 &\leq& k_2 - k_1 &\leq& k_2 &\leq& b \end{eqnarray} $$ because subtracting a positive number can only make a number smaller, never bigger. |
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Let's look at some example. We know that $$ 4<8, 2<10, $$ but we don't have $4-2<8-10$. On the other hand, $$ 3<9,4<8$$ but you cannot say $3-4>9-8$. There is nothing to deduce from substrating inequalities that are of the same side |
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No. The problem is that if $k_2\le b$, then $-k_2\ge-b$ |
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For an accurate proof that you cannot say anything about it: Suppose you would be able to find that $k_1-k_2\leq f(a,b)$ holds for some $f$, and $f(a,b)\in\mathbb{R}$. Then $k_2\geq k_1-f(a,b)$. And because $k_2\leq b$ we have $k_1-f(a,b)\leq b$. However, we can substitute $k_2=x=k_1-f(a,b)-1$. This value is allowed since $x\leq b$. But $x$ does not satisfy the condition $x\geq k_1-f(a,b)$. That means that $f(a,b)\not\in\mathbb{R}$, so $f(a,b)=\infty$ It may look like an otiose proof, but i don't think it is. |
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