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

Solution 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.

Solution 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.

Solution 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 $(a-b)(\ln a-\ln b)\geq 0$.

Solution 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 $a<b$