Prove that $x^{y^x} > y^{x^y}$ for $x > y > 1$.
So I've tried this so far:
$x^{y^x} > y^{x^y}$
$e^{(y^x)\ln x} > e^{(x^y)\ln y}$
$(y^x)\ln x > (x^y)\ln y$
$e^{\ln({(y^x)\ln x)}} > e^{\ln({(x^y)\ln y})}$
$x\ln(y) + \ln(\ln x) > y\ln x + \ln(\ln y)$
And well, I got stuck there. Is there something I'm doing wrong or should I try a different approach?
Edit:
$x\ln(y) - y\ln(x) > \ln(\ln y)-\ln(\ln x) $
$x\ln(y) - y\ln(x) > \ln{\left(\frac{\ln y}{\ln x}\right)}$
$e^{x\ln y - y\ln x} > \frac{\ln y}{\ln x}$
Since $\frac{\ln y}{\ln x} < 1$, it would suffice to show that $\frac{e^{x\ln(y)}}{e^{y\ln(x)}} > 1$. So:
$x\ln y> y\ln x$
$\frac{x}{y} > \frac{\ln x}{\ln y}$
How do I prove the last bit though?
Ignore everything in the "Edit" section. We have:
(1) $x\ln(y) + \ln(\ln x) > y\ln x + \ln(\ln y)$
Let $u = \frac{\ln x}{\ln y} > 1$ and $v = \ln y > 0$
Then $x = e^{uv}$, $y = e^v$
By substituting x and y in (1) we get:
$e^{uv}v + \ln u + \ln v > e^vuv + \ln v$
$e^{uv}v + \ln u - e^vuv > 0$
$\ln u > v(ue^v - e^{uv})$
If $ue^v - e^{uv} \leq 0$ then ok.
Else let $ue^v - e^{uv} > 0$
$ue^{v} > e^{uv}$
$\ln u + v > uv$
$\ln u > v(u - 1)$
Now it will suffice to show that:
(2) $u - 1 > ue^v - e^{uv}$
Let $f(v) = u - 1 - ue^v + e^{uv}$, $v \geq 0$
Since $f(0) = 0$, we can now prove that $f'(v) > 0$ for $v > 0$.
$f'(v) = ue^{uv} - ue^v > 0$ as $u > 1$
Which means that f is increasing for $v > 0$ and therefore (2) is satisfied.