# Inequality proof

1. Mar 22, 2010

### Hobold

Sorry for not being very descriptive in the title, but here's my problem:

Prove that for all $$a,b,c,d \ge 0$$, the following inequality is valid:

$$\frac{a+b+c+d}{4} \ge \sqrt[4]{abcd}$$

I've tried to make it this way:

$$(x-y)^2 \ge 0$$
$$x^2 + y^2 \ge 2xy$$

So:

$$[(a+b)-(c+d)]^4 \ge 0$$
$$[ (a+b)^2 - 2(a+b)(c+d) + (c+d)^2 ] ^2 \ge 0$$
$$(a+b)^2 + (c+d)^2 \ge 2(a+b)(c+d)$$
$$(a+b)^2 + 2(a+b)(c+d) + (c+d)^2 \ge 4(a+b)(c+d)$$
$$a+b+c+d \ge 2 \sqrt{(a+b)(c+d)}$$
$$\frac{a+b+c+d}{2} \ge \sqrt{(a+b)(c+d)}$$

This is the farthest I got, can't figure out how to continue... any tips?

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Another:

Prove, by induction, that the following inequality is true:

$$\prod_{n=1} \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}$$

$$\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{1}{\sqrt{2n+1}} \cdot \frac{2n}{2n+1}$$

$$\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)(2n+1)^2}}$$

$$\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)^3}}$$

Though I am really new to induction, so I'm not sure if my proof is finished.

Last edited: Mar 22, 2010
2. Mar 22, 2010

### Office_Shredder

Staff Emeritus
The notation for your induction proof is terrible. You should be doing something like multiplying from n=1 to N, and then adding on $$\frac{2N}{2N+1}$$ at the end. As written, the left hand side of each of those equations doesn't really make any sense (it's like saying $$(\sum_{i=0}i^2) + (i+1)^2$$... i is a dummy variable, you can't use it outside the summation like this. What you really want to write in this case is $$(\sum_{i=0}^n i^2) + (n+1)^2$$

3. Mar 22, 2010

### Hobold

Ok, I tried to abbreviate the notation, but apparently I failed miserably.

It's

$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}$$

4. Mar 22, 2010

### JSuarez

Do you know the Arithmetic-Geometric Inequality?

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means" [Broken]

Last edited by a moderator: May 4, 2017
5. Apr 5, 2010

### rs1n

From $x^2+y^2 \ge 2xy$, consider adding $2xy$ to both sides of the inequality. You should quickly see that

$$\frac{x+y}{2} > \sqrt{xy}$$

Now try apply the above to $x=\sqrt{ab}$ and $y=\sqrt{cd}$.

6. Apr 6, 2010

### IB1

Yeah this is only AM-GM for four numbers, but you can also use Holder's inequality or Jensen's inequality. In fact, AM-GM inequality is just a corollary of Jensen's inequality (hint- for proof consider the function f(x)=ln x). As for AM-GM inequality, it is:
If $$a_i , i=1,2,...,n$$ are positive real numbers then:
$$\frac{a_1 +a_2 +...+a_n}{n} \geq \sqrt[n]{a_1 a_2 ...a_n}$$
This can be generalized into something that we call General Mean Inequality.

Anyways, To prove this you can use induction on $$n$$. However, to prove the inequality that you posted, you may directly see that if the inequality holds for $$n=2$$ (for which you have the proof) then obviously holds for every $$n$$ which is a power of $$2$$.

As for you second inequality, take the $$ln$$ of both sides, and then try to use Jensen's inequality (hint- what can you say about the second derivative of f(x)=ln x ? )

7. Apr 6, 2010

### rs1n

Holder's and Jensen's inequalities will also do it, but it is much easier to understand using only elementary arguments. Just apply

$$\frac{x+y}{2} > \sqrt{xy}$$

sufficiently many times (namely with $x=\sqrt{ab}$ and $y=\sqrt{cd}$). What is $\sqrt{xy}$ in terms of a, b, c, and d? What is $\frac{x+y}{2}$ in terms of a, b, c, and d? Finally, consider applying the inequality above for $\sqrt{ab}$ (and similarly for $\sqrt{cd}$). This will not only give you the desired inequality, it will also give you insight into the inductive step of the general inequality.

Last edited: Apr 6, 2010
8. Apr 6, 2010

### IB1

@rs1n: Can you post/direct your proof of General Mean Inequality using induction.

Personally, I am aware of only one proof for it, it is simply using the fact that:
$$f(n)=\left(\frac{a_1 ^n +a_2 ^n +...+a_n ^n}{n}\right)^{1/n}$$ is increasing function, for $$a_i$$ being positive.

Last edited: Apr 6, 2010
9. Apr 6, 2010

### rs1n

Sorry, my mind must have fixated on the phrase "general mean inequality" when I actually meant general inequality -- as in for any n, not just n=4 -- in my response. As for the GMI, the only proof I know is likely the same as that one you are familiar with.

10. Apr 6, 2010

### Tobias Funke

The original (I believe) proof was by induction, but starting with powers of 2 and then filling in the rest. That proof is right on the wikipedia page for AM-GM.

edit: Oops, looks like I can't read. I don't know if this technique can be used for the more general inequality you guys are talking about.

Last edited: Apr 6, 2010