For $a,b,c$ be non-negative reals such that $a\leq b+c$. Then prove that there exists non-negative reals $x$ and $y$ such that $x+y=a$, $x\leq b$ and $y \leq$ c.
$a\geq 0, b\geq 0, c \geq 0$ and $a\leq b+c$, then there exists $x\geq 0 $ and $y\geq 0$ such that $x+y=a$, $x\leq b$ and $y \leq c $.
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real-analysis
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0I have taken cases where $b$ and $c$ are lower than $a$, both greater than $a$ and one on each side. No idea to proceed. – 2017-01-21
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If either $b\geq a$ or $c\geq a$, then $0+a=a$ works. If both $b$ and $c$ are smaller than $a$, then see that $a=b+(a-b)$, and $x=b\leq b$ and $0\leq y=a-b\leq c$.
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0Thanks a lot Arthur. Its done. – 2017-01-21