Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?

share|improve this question
    
What is $\Bbb N^*$? –  David Mitra Nov 9 '11 at 15:43
    
The set of natural numbers without the number 0 –  nick Nov 9 '11 at 15:45
    
I don't think this is number theory :( –  user9413 Nov 9 '11 at 16:04

4 Answers 4

up vote 3 down vote accepted

Here is a much more elementary proof:

$$a^{3a}b^{3b}=a^3a^3 \cdot... a^3 b^3b^3 \cdot ....b^3 \,.$$

Using the AM-GM inequality with $x_1=...=x_a=a^3$ and $x_{a+1}=...=x_{a+b}=b$ Yields

$$\sqrt[a+b]{a^3a^3 \cdot ... a^3 b^3b^3 \cdot ....b^3} \leq \frac{aa^3+bb^3}{a+b} \,.$$

Thus

$$a^{3a}b^{3b} \leq \left( \frac{a^4+b^4}{a+b} \right)^{a+b} \,.$$

share|improve this answer

Since $\log(x)$ is concave, $$ \log\left(\frac{ax+by}{a+b}\right)\ge\frac{a\log(x)+b\log(y)}{a+b}\tag{1} $$ Rearranging $(1)$ and exponentiating yields $$ \left(\frac{ax+by}{a+b}\right)^{a+b}\ge x^ay^b\tag{2} $$ Plugging $x=a^3$ and $y=b^3$ into $(2)$ gives $$ \left(\frac{a^4+b^4}{a+b}\right)^{a+b}\ge a^{3a}b^{3b}\tag{3} $$ and $(3)$ is the cube of the posited inequality.

From my comment (not using concavity):

For $0<t<1$, the minimum of $t+(1-t)u-u^{1-t}$ occurs when $(1-t)-(1-t)u^{-t}=0$; that is, when $u=1$. Therefore, $t+(1-t)u-u^{1-t}\ge0$. If we set $u=\frac{y}{x}$ and $t=\frac{a}{a+b}$, we get $$ \frac{ax+by}{a+b}\ge x^{a/(a+b)}y^{b/(a+b)}\tag{4} $$ Inequality $(2)$ is simply $(4)$ raised to the $a+b$ power.

share|improve this answer
    
I appreciate the effort, however, I don't know about concave functions. Does there exist a simpler solution? –  nick Nov 9 '11 at 16:32
    
@nick: I think most other approaches will be more complex. However, note that for $0<t<1$, the minimum of $t+(1-t)u-u^{1-t}$ occurs when $(1-t)-(1-t)u^{-t}=0$, that is $u=1$. Therefore, $t+(1-t)u-u^{1-t}\ge0$. Let $u=y/x$ and $t=a/(a+b)$ and we get $$\frac{ax+by}{a+b}\ge x^{a/(a+b)}y^{b/(a+b)}$$ Raise to the $a+b$ power and you have $(2)$. –  robjohn Nov 9 '11 at 17:10
    
@nick: I've moved the last comment into the answer. –  robjohn Nov 9 '11 at 19:18

This is expanding on a comment by Bill, the following might work:

You need

$$ (a+b)\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq a \ln(a) + b \ln(b) \,.$$

Or

$$ (\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq \frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b) \,.$$

Now, if I remember right, the Jensen inequality for Log reads:

$$\frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b) \leq \ln (\frac{a^2+b^2}{a+b}) \,.$$

Thus, you only need to show

$$\left( \frac{a^2+b^2}{a+b} \right)^3 \leq \frac{a^4+b^4}{a+b} \,.$$

Or

$$(a^2+b^2)^3 \leq (a+b)^2(a^4+b^4) \,.$$

EDIT After a long calculation, this reduces to

$$a^6+3a^4b^2+3a^2b^4+b^6 \leq a^6+a^2b^4+2a^5b+2ab^5+a^2b^4+b^6$$

or

$$a^4b^2+a^2b^4 \leq a^5b+ab^5$$

After canceling $ab$ this follows imediatelly form the AM-GM.: $a^3b \leq \frac{a^4+a^4+a^4+b^4}{4}$ and $ab^3 \leq \frac{a^4+b^4+b^4+b^4}{4}$

share|improve this answer
    
Can you show it without Jensen's inequality and by using the fact that a,b are natural numbers? –  nick Nov 9 '11 at 16:12
2  
Please don't use "foiling" without defining it. Non-native speakers or native speakers with a non-US education cannot look it up in a dictionary. –  Phira Nov 9 '11 at 16:18

With $p=a/(a+b)$ and $q=b/(a+b)$ you can use to homogeneity to get

$$\begin{align} &&(a^4+b^4)^{a+b}&\ge (a+b)^{a+b}a^{3a}b^{3b} \\&\Leftrightarrow& (a^4+b^4)&\ge (a+b)a^{3p}b^{3q} \\ &\Leftrightarrow &p^4+q^4&\ge p^{3p}q^{3q}\\ & \Leftrightarrow & \sqrt[3]{p \cdot p^3 + q \cdot q^3} &\ge p^p q^q, \end{align}$$

which is exactly the weighted mean inequality between the cubic mean and the geometric mean of $p$ and $q$ with weights $p$ and $q$.

(Obviously, the proofs of the general mean inequality are similar to the other posted answers, but one does not have to repeat the proof for each instance.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.