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

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

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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

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

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

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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.

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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 $ab^3 \leq \frac{a^4+b^4+b^4+b^4}{4}$

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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
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.)

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