$$ x \geq y $$ $$ a \geq b $$

$$x+a \geq y+b $$ is valid but $$ x-a \geq y-b $$ is not valid

Can we say the latter is valid if $x-a \geq 0$ ?

Is it a proof or am I wrong? Are there counter examples?

If $x,y,a,b \geq 0 $ ?


2 Answers 2


If $a \geq b$ then $-a \leq -b$. So you could for sure deduce that $$x-b \geq y-a$$ As for a counter example in the light of the assumption $x-a \geq 0$, try $x = y = a = 1$ with $b = 0$.

  • $\begingroup$ What if x,y,a,b >0 ? $\endgroup$
    – S.Dan
    Jun 14, 2015 at 17:07
  • 1
    $\begingroup$ @S.Dan Take $x = y = a = 2$ with $b = 1$ $\endgroup$
    – graydad
    Jun 14, 2015 at 17:08

Your inequality is only true when the difference between $x$ and $y$ is bigger than the difference between $a$ and $b$. That is,

$$x-y \geq a-b$$

However, that is just a rewrite of the inequality you gave. $x\geq y$ and $a\geq b$ don't say anything about $x-a$ and $y-b$.

So to find a counterexample, one could choose $x=3$, $y=5$, $a=2$, $b=9$ and get $1 > 4$, which is not true.


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