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Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.

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    $\begingroup$ I am using this post to examine the possible usage (abuses) of bounties as an "investment" for reputations. For example, if you have a really good solution, then put a 50 pt. bounty to it, most probably more than 5 upvotes will be given $\endgroup$ – user198454 Dec 13 '14 at 14:37
  • $\begingroup$ @TheMathTroll Alas, one must be way of the 200 rep daily upvote cap! $\endgroup$ – Emily Dec 16 '14 at 22:29
  • $\begingroup$ Finally got the most generalized form $a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_1^{a_1} \ge a_1^{a_2}a_2^{a_3} \cdots a_{n-1}^{a_n }a_n^{a_1}$ $\endgroup$ – Shivam Patel Dec 18 '14 at 10:48
  • $\begingroup$ post it somewhere $\endgroup$ – user198454 Dec 18 '14 at 13:59
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    $\begingroup$ Apparently the way to get your question voted up a lot is to put 'Oxford' in the title. This is absurd. $\endgroup$ – punctured dusk Dec 23 '14 at 10:46

10 Answers 10

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We just have to show that $a^{a-b} \ge b^{a-b}$. This is equivalent to $(\frac{a}{b})^{a-b} \ge 1$.

If $a \ge b$, then $\frac{a}{b} \ge 1$, Also $a-b \ge 0$. A number greater than $1$ raised to a positive exponent is clearly greater than $1$.

If $a \le b$, then $\frac{a}{b}\leq 1$. $a-b\leq 0$. A positive number less than $1$ raised to a negative exponent is greater than $1$.

Hence we are done as we considered both cases.

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    $\begingroup$ I would think since there is symmetry under $a \mapsto b$ we would only need to consider one case. $\endgroup$ – Bennett Gardiner Dec 12 '14 at 9:59
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    $\begingroup$ 29 upvotes... how's that possible. I actually thought this was relatively poorly-written and would receive tens of downvotes... Yeah, symmetry is nice. This is very similar to the rearrangement inequality. $\endgroup$ – user198454 Dec 12 '14 at 12:05
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    $\begingroup$ The Math Troll is gracious! Not such a troll at all, in fact... :O) $\endgroup$ – Ellie Kesselman Dec 12 '14 at 15:07
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    $\begingroup$ @TheMathTroll "Gracious" as in appreciative, modest and endearing. Don't make me change my mind! I was just trying to be nice. I meant no offense. I was trying to complement you, not enrage you. $\endgroup$ – Ellie Kesselman Dec 12 '14 at 22:55
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    $\begingroup$ I am using this post to examine the possible usage (abuses) of bounties as an "investment" for reputations. For example, if you have a really good solution, then put a 50 pt. bounty to it, most probably more than 5 upvotes will be given $\endgroup$ – user198454 Dec 13 '14 at 14:25
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$$\log(a^a b^b)=a \log a + b \log b$$ $$\log(a^b b^a)=a \log b + b \log a$$ Thus, by the rearrangement inequality, because $\log$ is strictly increasing, $$\log(a^a b^b)\geq \log(a^b b^a)$$ Similarly, because $\log$ is strictly increasing, $$a^a b^b \geq a^b b^a.$$

This can be generalized as follows. Let $\sigma_1, \sigma_2, ..., \sigma_n$ be any permutation of $1, 2, ..., n$, then $$ a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_n^{a_n} \ge a_1^{a_{\sigma_1}}a_2^{a_{\sigma_2}} \cdots a_{n-1}^{a_{\sigma_{n-1}}}a_n^{a_{\sigma_n}} \ge a_1^{a_n} a_2^{a_{n-1}}\cdots a_{n-1}^{a_2} a_n^{a_1} $$ by an identical argument.

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  • $\begingroup$ What if $\log a$ or $\log b\leq0$? Better assume $a\geq b$, and use $\log_{b/2}$. $\endgroup$ – A. Chu Dec 12 '14 at 3:54
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    $\begingroup$ @ᴊᴀsᴏɴ $\log x$ is strictly increasing for all values $x>0$, and the rearrangement inequality applies for both negative and positive numbers. $\endgroup$ – Suzu Hirose Dec 12 '14 at 3:58
  • $\begingroup$ Ah, i see, i thought it only apply to positive reals. $\endgroup$ – A. Chu Dec 12 '14 at 3:58
  • $\begingroup$ This is neat. It seems to be the most natural and intuitive way to show the result. $\endgroup$ – user795305 Dec 17 '14 at 14:46
  • $\begingroup$ Great solution. I wonder if somebody could make another, unrelated, generalization. $\endgroup$ – VividD Dec 31 '14 at 17:49
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This inequality is equivalent to $a\ln a+b\ln b\geq a\ln b+b\ln a$, which is obvious once rearranged as $(a-b)(\ln a-\ln b)\geq 0$.

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$$ \left(\frac ab\right)^{a-b}-1=\frac{a^ab^b-a^bb^a}{b^aa^b}$$

If $a=b, \left(\dfrac ab\right)^{a-b}=1$

Else if $a>b;\dfrac ab>1$ and $a-b>0\implies \left(\dfrac ab\right)^{a-b}>1$

Similarly if $a<b$

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$$a^a \ b^b \;?\; a^b \ b^a \\ \frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\ \left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\ \left(\frac{a}{b}\right)^{a-b} \;?\; 1 $$

if $a \ge b$, then $c = \frac{a}{b} \ge 1$, and $d = a-b \ge 0$. Thus $c^d \ge 1$, so $?$ is $\ge$.

if $a < b$, then:

$$\left(\frac{a}{b}\right)^{a-b} \\ = \left(\frac{b}{a}\right)^{b-a}$$

and we re-use the first result by symmetry.

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  • $\begingroup$ I like the question mark notation, very useful when one has to decide which of two things is the biggest :) $\endgroup$ – punctured dusk Dec 17 '14 at 16:35
  • $\begingroup$ @barto: Thanks, it made sense to me, but isn't something I've seen done in literature. Perhaps it goes without saying, but because ? could be an inequality, the same care must be taken about multiplying by negative or possibly negative numbers. Maybe an upside-down question mark when reversing the inequality? $\endgroup$ – Phil H Dec 22 '14 at 10:18
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Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$.

$b$ can be expressed as $a+x$, where $x$ is some positive integer.

to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$

Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$

$= a^a (a+x)^{a+x} - a^{a+x} (a+x)^a$

expanding the powers

$= a^a (a+x)^a (a+x)^x - a^a a^x (a+x)^a$

taking common factors out of bracket, we get

$= a^a (a+x)^a [ (a+x)^x - a^x ] $

since $x \gt0$, $(a+x)^x$ must be greater than $a^x$, therefore the above expression evaluates to a value greater than zero.

The other case of $a\gt b$ is the same as this since the order of $a$ and $b$ don't matter, while $a=b$ is trivial.

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Got the full generalization for $a_1,a_2, \cdots a_n$ We have $$a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_1^{a_1} \ge a_1^{a_2}a_2^{a_3} \cdots a_{n-1}^{a_n }a_n^{a_1}$$ This can be proved using the fact that if $x >0$, then $f(x)=x^{1/x}$ has absolute maximum at $e$ (Euler constant) .Thus for any positive real number $x$ $$e^x \ge x^e$$
We begin our proof by substituting $x=\frac{a_1e}{a_2}$ , $x=\frac{a_2e}{a_3}$ $\cdots$ $x=\frac{a_ne}{a_1}$. Thus $$e^{a_1e} \ge (\frac{a_1e}{a_2})^{a_2e} , \quad e^{a_2e} \ge (\frac{a_2e}{a_3})^{a_3e} \cdots \quad e^{a_ne} \ge (\frac{a_ne}{a_1})^{a_1e}$$ Multiplying these gives $$e^{(a_1+a_2+ \cdots + a_n)e} \ge (\frac{a_1}{a_2})^{a_2e} (\frac{a_2}{a_3})^{a_3e} \cdots(\frac{a_n}{a_1})^{a_1e}e^{(a_1+a_2+\cdots + a_n)e}$$ Hence $$1 \ge(\frac{a_1}{a_2})^{a_2e} (\frac{a_2}{a_3})^{a_3e} \cdots(\frac{a_n}{a_1})^{a_1e}$$ Which implies $$1 \ge (\frac{a_1}{a_2})^{a_2} (\frac{a_2}{a_3})^{a_3} \cdots(\frac{a_n}{a_1})^{a_1}$$ Which finally leads to $$a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_1^{a_1} \ge a_1^{a_2}a_2^{a_3} \cdots a_{n-1}^{a_n }a_n^{a_1}$$

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    $\begingroup$ $a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_n^{a_n} \ge a_1^{a_{\sigma_1}}a_2^{a_{\sigma_2}} \cdots a_{n-1}^{a_{\sigma_{n-1}}}a_n^{a_{\sigma_n}}$ for any permutation $\sigma$ drops straight out of the rearrangement inequality by the identical argument to that shown in my post. $\endgroup$ – Suzu Hirose Dec 18 '14 at 12:40
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We'll show a bit more: $a^a b^b > \left( \frac{a+b}{2} \right)^{a+b} > a^b b^a.$

First we need a lemma: $P = (1+x)^{1+x} (1-x)^{1-x} > 1$ if $x < 1$.

Proof: $\ln P = (1+x)\ln (1+x) + (1-x) \ln (1-x) = x ( \ln (1+x) - \ln (1-x)) + \ln (1+x) + \ln (1-x) = 2x \left(x + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right) - 2 \left( \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \cdots \right) = 2 \left(\frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \cdots \right).$

Therefore, $\ln P$ is positive, and so $P > 1.$

In this lemma put $x = \frac{s}{t}$, where $t > s$. Then $ \left( 1 + \frac{s}{t} \right)^{1+ \frac{s}{t}} \left(1 - \frac{s}{t}\right)^{1-\frac{s}{t}} > 1. $ Thus $ \left( \frac{t+s}{t} \right)^{t+s} \left(\frac{t-s}{t} \right)^{t-s} > 1^t = 1.$ Thus $ (t + s)^{t+s}(t - s)^{t-s} > t^{2t}$. Letting $t + s = a$ and $t - s = b$, we see that $t = \frac{a+b}{2}$ and so $a^a b^b > \left( \frac{a+b}{2}\right)^{a+b}.$

In a similar way, letting $Q = (1+x)^{1-x} (1-x)^{1+x}$, we can show that $\ln Q < 0$ and so $Q < 1.$ Now continue as we did in the last proof.

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  • $\begingroup$ Also, for positive real numbers $a, b, c$ we have $\endgroup$ – PolyaPal Dec 17 '14 at 15:35
  • $\begingroup$ It is known that $a^a b^b \ge (ab)^{(a+b)/2}$. Also, for positive real numbers $a, b, c$ we have $a^a b^b c^c \ge \left(\frac{a+b+c}{3}\right)^{a+b+c} \ge(abc)^{(a + b + c)/3}$. $\endgroup$ – PolyaPal Dec 17 '14 at 15:41
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There are three cases: $a = b$, $a \lt b$, and $a \gt b$. The inequality is trivial when $a = b$, and the case where $a \lt b$ is really the same as the case where $a \gt b$. So lets just consider the case where $a \lt b$ and rewrite the the equation a little:

(1) $a^a \ b^b \ge a^b \ b^a$

with my assumption can be rewritten as

(2) $a^a \ b^a \ b^{b-a} \ge a^a \ b^a \ a^{b-a}$

Canceling the common terms gives

(3) $b^{b-a} \ge a^{b-a}$

That's it really, because since $b \gt a$, a bunch of $b$s multiplied together are greater than a bunch of $a$s multiplied together ("bunch" = $b-a$)

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$$ a^a b^b \overset{?}{\ge} a^b b^a \\ (a/b)^a \overset{?}{\ge} (a/b)^b $$

Now if $a \gt b$ then this is obviously true, because the parenthese evaluates to some number $x$ with $1 \lt x$ and such a number to a higher power (on the lhs) is greater than that of a smaller power (on the rhs).

But if $b \gt a$ then this is again obviously true, because the parenthese evaluates to $0 \lt x \lt 1 $ and such a number to a higher power (on the rhs) is smaller than that of a smaller power (at the lhs).

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