When is $a+b\leq ab$, a and b are positive real numbers
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Is this true!
Given $a,b>0$, real numbers, then $$a+b\leq ab$$ If not, when this could be true?
algebra-precalculusinequality
asked 2011-10-29
user id:18476
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Clearly, if $a=1$ or $b=1$, it can't be true. You then check the cases of, say, $a > 1$ and $a < 1$... – 2011-10-29
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Just an interesting comment. This can be extended to complex numbers too. See $(1+i)+(1-i)=(1+i)(1-i)$ and $2=2$. – 2011-10-29
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$(a_1+a_2i)+(b_1+b_2i) \leq (a_1+a_2i)(b_1+b_2i)$$(a_1+b_1)+(a_2+b_2)i \leq (a_1b_1-a_2b_2)+(a_1b_2+a_2b_1)i$$(a_1+b_1-a_1b_1+a_2b_2)+(a_2+b_2-a_1b_2-a_2b_1)i \leq 0$ But, this inequation just holds if $(a_2+b_2-a_1b_2-a_2b_1)=0$ and we have too $(a_1+b_1-a_1b_1+a_2b_2) \leq 0$ – 2011-10-29