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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0I don't think this is number theory :( – 2011-11-09
4 Answers
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.)
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
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0@nick: I've moved the last comment into the answer. – 2011-11-09
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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2Please 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. – 2011-11-09
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
$a^{3a}b^{3b} \leq \left( \frac{a^4+b^4}{a+b} \right)^{a+b} \,.$