# proving inequality?

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 $1/a\mapsto a^2$. – Generic Human Jul 2 '12 at 16:39
@Generic Human: Thanks for the hint. It worked out. Nice trick! – C. Lambda Jul 2 '12 at 16:47

$\frac {bc}{a}+\frac{ca}{b}+\frac{ab}{c}=\frac{1}{2}(\frac{ca}{b}+\frac{ab}{c})+\frac{1}{2}(\frac{ab}{c}+\frac{bc}{a})+\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b}\geq\sqrt\frac{ca}{b}\sqrt\frac{ab}{c}+\sqrt\frac{ab}{c}\sqrt\frac{bc}{a}+\sqrt\frac{bc}{a}\sqrt\frac{ca}{b}=a+b+c$
Apply your first inequality to $a=\frac 1 x, b=\frac 1 y, c=\frac1z$ and clear fractions.
If $a<b$ then it is only true that $ca<cb$ if $c$ is positive. If $c=0$ we have equality and if $c<0$ the inequality reverses. – Mark Bennet Jul 2 '12 at 18:24