Proving Inequalities to Induction and Simplification

[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}}

I made this:

\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.

 

[Office_Shredder]

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

 

[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}}

 

[JSuarez]

Do you know the Arithmetic-Geometric Inequality?

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means"

 

[rs1n]

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

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}.

 

[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 ? )

 

[rs1n]

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 ? )

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

<br /> \frac{x+y}{2} &gt; \sqrt{xy}<br />

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.

 

[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.

 

[rs1n]

@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)^n is increasing function, for a_i being positive.

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.

 

[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.