An interesting inequality-question on the proof

1. Sep 4, 2008

jeffreydk

An interesting inequality--question on the proof...

I am working on the following proof and have gotten about half way;

If $$a_1, a_2, \ldots , a_n$$ are positive real numbers then

$$\sqrt[n]{a_1 \cdots a_n} \leq \frac{a_1+a_2+\ldots +a_n}{n}$$

By induction, I started by showing it for n=2, which goes as follows.

We must show that $$\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}$$. So consider some s, a positive real number. If

$$\sqrt{a_1 a_2} \leq s \Rightarrow \sqrt{a_1 a_2} \leq \frac{\sqrt{a_1a_2}+s}{2} \leq s$$

And since $$\sqrt{a_1a_2}<a_1+a_2-\sqrt{a_1a_2}$$, just let that equal s. Then it follows that,

$$\sqrt{a_1a_2} \leq \frac{\sqrt{a_1a_2}+a_1+a_2-\sqrt{a_1a_2}}{2}$$

Where after simplification we have shown that $$\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}$$ or in other words it is true for n=2.

Now for the next part of the proof, I think we must show that if it is true for $$n=2^m$$ then it is also true for $$n=2^{m+1}$$. And then we must also show it if $$2^m < n < 2^{m+1}$$, but this is where I am stumped. Sorry for the choppiness in my explanation. Thanks for any assistance, I appreciate it.

2. Sep 4, 2008

snipez90

Re: An interesting inequality--question on the proof...

Eh, well the n = 2 case could be proved in an easier way just be multiplying both side by 2, squaring, rearranging, and factoring. It follows from the trivial inequality $$x^2 \geq 0$$

As for the induction, I'm not sure why you use n = 2^m since it works for any natural number n. Anyways AM-GM is not easy to prove by induction since you have to apply an algorithm that is also used in a proof without induction.

3. Sep 5, 2008

morphism

Re: An interesting inequality--question on the proof...

Actually, one of the classical proofs of the AM-GM inequality is an inductive one. The proof is due to Cauchy, and it follows the main ideas in the OP, i.e. inducting on powers of 2.

There was a thread on this sometime back, here is the link. Check it out and post back if you need further help.

4. Sep 5, 2008

jeffreydk

Re: An interesting inequality--question on the proof...

Oh alright, I didn't realize this was such a famous proof (Cauchy). Thanks for alerting me to that reference.

5. Sep 6, 2008

Count Iblis

Re: An interesting inequality--question on the proof...

It can also be derived from the inequality

$$\exp(x)\geq 1+x$$ (1)

This is trivial to prove. It follows from this that:

$$\left\langle\exp(X)\right\rangle = \exp\left(\left\langle X\right \rangle\right)\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle$$

Apply (1) to the last factor:

$$\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle\geq \left\langle 1 + X - \left\langle X\right \rangle\right\rangle = 1$$

So, we have:

$$\left\langle\exp(X)\right\rangle \geq \exp\left(\left\langle X\right \rangle\right)$$ (2)

This is a special case of the convex inequality and the AM-GM inequality is a straightforward consequence of that inequality. But the convex inequality requires a bit more work to prove. The AM-GM inequality follows from (2) by taking X to be the logarithms of the a_i. The average on the l.h.s. is then the arithmetic mean, while on the r.h.s. you get the geometric mean.

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An interesting inequality-question on the proof

1. Sep 4, 2008

jeffreydk

An interesting inequality--question on the proof...

I am working on the following proof and have gotten about half way;

If $$a_1, a_2, \ldots , a_n$$ are positive real numbers then

$$\sqrt[n]{a_1 \cdots a_n} \leq \frac{a_1+a_2+\ldots +a_n}{n}$$

By induction, I started by showing it for n=2, which goes as follows.

We must show that $$\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}$$. So consider some s, a positive real number. If

$$\sqrt{a_1 a_2} \leq s \Rightarrow \sqrt{a_1 a_2} \leq \frac{\sqrt{a_1a_2}+s}{2} \leq s$$

And since $$\sqrt{a_1a_2}<a_1+a_2-\sqrt{a_1a_2}$$, just let that equal s. Then it follows that,

$$\sqrt{a_1a_2} \leq \frac{\sqrt{a_1a_2}+a_1+a_2-\sqrt{a_1a_2}}{2}$$

Where after simplification we have shown that $$\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}$$ or in other words it is true for n=2.

Now for the next part of the proof, I think we must show that if it is true for $$n=2^m$$ then it is also true for $$n=2^{m+1}$$. And then we must also show it if $$2^m < n < 2^{m+1}$$, but this is where I am stumped. Sorry for the choppiness in my explanation. Thanks for any assistance, I appreciate it.

2. Sep 4, 2008

snipez90

Re: An interesting inequality--question on the proof...

Eh, well the n = 2 case could be proved in an easier way just be multiplying both side by 2, squaring, rearranging, and factoring. It follows from the trivial inequality $$x^2 \geq 0$$

As for the induction, I'm not sure why you use n = 2^m since it works for any natural number n. Anyways AM-GM is not easy to prove by induction since you have to apply an algorithm that is also used in a proof without induction.

3. Sep 5, 2008

morphism

Re: An interesting inequality--question on the proof...

Actually, one of the classical proofs of the AM-GM inequality is an inductive one. The proof is due to Cauchy, and it follows the main ideas in the OP, i.e. inducting on powers of 2.

There was a thread on this sometime back, here is the link. Check it out and post back if you need further help.

4. Sep 5, 2008

jeffreydk

Re: An interesting inequality--question on the proof...

Oh alright, I didn't realize this was such a famous proof (Cauchy). Thanks for alerting me to that reference.

5. Sep 6, 2008

Count Iblis

Re: An interesting inequality--question on the proof...

It can also be derived from the inequality

$$\exp(x)\geq 1+x$$ (1)

This is trivial to prove. It follows from this that:

$$\left\langle\exp(X)\right\rangle = \exp\left(\left\langle X\right \rangle\right)\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle$$

Apply (1) to the last factor:

$$\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle\geq \left\langle 1 + X - \left\langle X\right \rangle\right\rangle = 1$$

So, we have:

$$\left\langle\exp(X)\right\rangle \geq \exp\left(\left\langle X\right \rangle\right)$$ (2)

This is a special case of the convex inequality and the AM-GM inequality is a straightforward consequence of that inequality. But the convex inequality requires a bit more work to prove. The AM-GM inequality follows from (2) by taking X to be the logarithms of the a_i. The average on the l.h.s. is then the arithmetic mean, while on the r.h.s. you get the geometric mean.

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The text passage contains a mathematical question in the form of a proof problem and multiple suggestions and discussions about how to solve it, but no direct answer to the mathematical question is provided.

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