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Here is another inequality I am trying to prove:

Let $a,b,c$ be positive numbers. Prove that:

$$1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$$ $$2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$$

In the book's hint, it uses the inequality: $$a^{2}+b^{2}+c^{2}\geq ab+bc+ca$$ (which is easy to prove), then it follows that : $$b^{2}c^{2}+a^{2}c^{2}+a^{2}b^{2}\geqslant abc(a+b+c)$$ which is equivalent to proving our claim. I need to know how the second inequality follows from the first one. Also, any suggestions for proving the second claim?

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    For the first claim, see what happens when mapping $a\mapsto 1/a$ (and so on for the other variables) and reducing to a common denominator. For the second claim, just eliminate the inverses and square roots by $1/a\mapsto a^2$.2012-07-02
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    @Generic Human: Thanks for the hint. It worked out. Nice trick!2012-07-02

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$a^2+b^2+c^2\geq ab+ac+bc$ apply $a\rightarrow bc, b\rightarrow ca, c\rightarrow ab$ we have $a^2b^2+b^2c^2+c^2a^2\geqslant a^2bc+b^2ca+c^2ab=abc(a+b+c)$

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    What does "apply$a\rightarrow bc, b\rightarrow ca, c\rightarrow ab"$ mean?2016-11-24
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    ok, understood, it means substituting $a$ by $bc$ and so on in a cyclical way2016-11-24
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    Yes. That's what i mean2016-11-25
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Apply your first inequality to $a=\frac 1 x, b=\frac 1 y, c=\frac1z$ and clear fractions.

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    Thanks for the answer. Indeed, one the substitution is done, everything follows immediately. Nice trick!!2012-07-02
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    There is a clue to this in the formulation of the question. The first inequality is true whether or not the numbers are positive - so where are you going to use that condition. Well, it would be necessary if your were going to multiply both sides of an inequality by a number ...2012-07-02
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    I didn't quite understand your last hint. what do you mean by multiplying both sides of the inequality by a number? Please elaborate2012-07-02
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    If $a then it is only true that $ca if $c$ is positive. If $c=0$ we have equality and if $c<0$ the inequality reverses.2012-07-02
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    Excuse me, but do we need this to solve the second part?2012-07-02
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    No, but the condition in the question that the numbers are positive is crucial in clearing fractions in the first part. It is a clue that you have to use positivity somewhere in the answer.2012-07-02
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    You're right. Thanks for pointing out this2012-07-02
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$\frac {bc}{a}+\frac{ca}{b}+\frac{ab}{c}=\frac{1}{2}(\frac{ca}{b}+\frac{ab}{c})+\frac{1}{2}(\frac{ab}{c}+\frac{bc}{a})+\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b}\geq\sqrt\frac{ca}{b}\sqrt\frac{ab}{c}+\sqrt\frac{ab}{c}\sqrt\frac{bc}{a}+\sqrt\frac{bc}{a}\sqrt\frac{ca}{b}=a+b+c$