If both $a,b>0$, then $a^ab^b \ge a^bb^a$

Question

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

Answer 1

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.

Answer 2

$$\log(a^a b^b)=a \log a + b \log b$$ $$\log(a^b b^a)=a \log b + b \log a$$ 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.

Answer 3

This inequality is equivalent to $a\ln a+b\ln b\geq a\ln b+b\ln a$, which is obvious once rearranged as $(a-b)(\ln a-\ln b)\geq 0$.

Answer 4

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

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 $a<b$

Answer 5

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

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.

Answer 6

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.

Answer 7

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 $$a_1^{a_1} a_2^{a_2} a_3^{a_3} \cdots a_1^{a_1} \ge a_1^{a_2}a_2^{a_3} \cdots a_{n-1}^{a_n }a_n^{a_1}$$

Answer 8

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.

Answer 9

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" = $b-a$)

Answer 10

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

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).