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confused about the demonstration of the following statement : let a, b, c $\in R^{*+}$ .

Demonstrate that: $a*b\geq1$ OR $a+b \leq \frac{1}{a}+\frac{1}{b}$

Demonstrate that ($a*b\geq1$ And $a+b \leq \frac{1}{a}+\frac{1}{b}$) if and only if $a*b=1$

Let's suppose that ($a*b*c\geq1$ And $a+b+c \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$) , demonstrate that none of a,b and c is equal to 1 and that one of them is less than 1.

SAM.Am
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  • What are you confused about? – Yagger Sep 06 '19 at 10:48
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    @SAM.Am Can you elaborate on why you are confused? If you don't provide any details or share or own attempts, most people will feel they are doing your homework for you... – PierreCarre Sep 06 '19 at 10:51
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    The first two propositions seem straightforward enough, but the third proposition appears to be false; take $a=b=c=1$ as a counterexample. It's also strange to say "let $a,b,c\in\mathbb R^+$" and then prove that none of $a,b,c$ are real; moreover if none are real how can you say that one of them is less than $1.$ Is the last proposition correctly stated? – David K Sep 06 '19 at 10:55
  • @PierreCarre you are right, I tried to solve the second, which I posted underneath, but I am still confused about the lack of information related to the first statement. – SAM.Am Sep 06 '19 at 18:50
  • @DavidK I translated literally the third question from a French Exercise: On suppose que abc>=1 et a+b+c >= 1/a +1/b +1/c ). Démontrer qu'aucun des Nombres a,b,c n'est égal à 1, et que l'un d'entre eux est inférieur à 1. – SAM.Am Sep 06 '19 at 18:50
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    You seem to have statement c wrong. First, the second inequality in the comment is different from the one in the question. Also, the comment says, "Show that none of them is equal to $1$," which is translated in the question as "none of them is real." – saulspatz Sep 06 '19 at 19:10
  • @saulspatz thanks for your comment, I update it. Sorry for the confusion. – SAM.Am Sep 06 '19 at 19:19
  • The question is still missing the “$+c$” from the comment, and I still don’t understand why $a=b=c=1$ isn’t a counterexample. My French is extremely poor, so I don’t know if there’s something else in the original that is still missing from the translation. – David K Sep 06 '19 at 19:55

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second demonstration solution

Demonstrate that ($a*b\geq1$ And $a+b \leq \frac{1}{a}+\frac{1}{b}$) if and only if $a*b=1$

$a*b\geq1$ => $a\geq \frac{1}{b}$ and $b\geq \frac{1}{a}$

so we have \begin{cases} a+b \geq \frac{1}{a}+\frac{1}{b} \\a+b\leq \frac{1}{a}+\frac{1}{b} \end{cases}

therefore $a+b = \frac{1}{a}+\frac{1}{b}$ $=> a+b = \frac{a+b}{ab}$ so $ab = 1$

SAM.Am
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    Also it’s not clear whether you considered the “if” direction of the proof, though that direction is easy. – David K Sep 06 '19 at 19:48
  • @DavidK you are right, it was a typo. I updated it. Thanks. ( for the "if". direction, do you mean I ought to start from that condition ?) – SAM.Am Sep 06 '19 at 21:18
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    For the "if" I would start, "Given $ab = 1.$ Then ..." and you can pretty much write the result right away. Not a big thing, just a formality. – David K Sep 07 '19 at 02:15
  • @DavidK. that was my humble translation from French language. (si et seulement si ) – SAM.Am Oct 10 '19 at 05:09