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?

• 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.

• Thanks for the answer. Indeed, one the substitution is done, everything follows immediately. Nice trick!! – C. Lambda Jul 2 '12 at 16:48
• 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 ... – Mark Bennet Jul 2 '12 at 16:54
• I didn't quite understand your last hint. what do you mean by multiplying both sides of the inequality by a number? Please elaborate – C. Lambda Jul 2 '12 at 18:21
• 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
• Excuse me, but do we need this to solve the second part? – C. Lambda Jul 2 '12 at 18:26

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

• What does "apply$a\rightarrow bc, b\rightarrow ca, c\rightarrow ab"$ mean? – vidyarthi Nov 24 '16 at 5:37
• ok, understood, it means substituting $a$ by $bc$ and so on in a cyclical way – vidyarthi Nov 24 '16 at 5:41
• Yes. That's what i mean – TropicalDog Nov 25 '16 at 10:56