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$$ \frac ab + \frac bc + \frac ca \geq a + b + c $$ where $abc = 1$

Using AM-GM I can get $$ \frac ab + \frac bc +\frac ca + ab + ac + bc \geq 2(a + b + c), $$ but I can't show $ab + bc + ac \leq a + b + c$ because it isn't true, nor is it with the inequality reversed. If the above inequality is true doesn't $ab + ac + bc \leq a + b +c$ need to be true?

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I think this will be easy when you substitute $a=\frac xy$, $b=\frac yz$ and $c=\frac zx$. Note that $abc=1$ is satisfied then. –  Ragnar Jul 13 at 11:09
    
Thanks yes that makes it much easier. But could you consider my question at the end also? Is there no way to get from the second inequality to the required one? –  user121591 Jul 13 at 11:21
    
Is $a,b,c$ positive? –  Sawarnik Jul 13 at 12:34
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up vote 6 down vote accepted

Using the substitution $a=\frac xy$, $b=\frac yz$ and $c=\frac zx$, we want to prove that $$ \sum_{cyc}\frac {xy}{z^2}\geq \frac xy+\frac yz+\frac zx $$ where we are taking cyclic sums. Using AM-GM, we find that $$ \frac{\frac{xy}{z^2}+\frac{xz}{y^2} +\frac{xz}{y^2}}3 \geq \sqrt[3]{\frac{x^3}{y^3}}=\frac xy $$ Applying this three times yields the required inequality.

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Thanks, but why don't we require ab + bc + ac =< a + b + c? Is this not the only way to get from my second inequality to the required one? –  user121591 Jul 13 at 11:38
    
When $A+B\leq C+D$ and $A\leq C$, you can't say anything about the relation between $B$ and $D$. When $A\geq C$, you can subtract the inequalities to get $B\leq D$, but that doesn't apply here. –  Ragnar Jul 13 at 11:41
    
We want to prove $A+B\leq C+D$. When we know $A\leq C$, it still may be the case that both are true, but $B\geq D$ always, for example take $A=1$, $C=4$, $B=3$ and $D=1$. –  Ragnar Jul 13 at 11:43

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