How can I prove this inequality with powers of radicals? The inequality goes like this:
$$\sqrt7^{\sqrt5}>\sqrt5^{\sqrt7}$$
I have to prove this without using approximate numbers and I just cannot find out how.
 A: This inequality is equivalent to
$$\sqrt 5\ln(\sqrt7) > \sqrt 7\ln(\sqrt5)\iff \frac{\ln(\sqrt7)}{\sqrt7}>\frac{\ln(\sqrt 5)}{\sqrt5}.$$
Consider the function $\;f(x)=\dfrac{\ln x}x$. As  $\;f'(x)=\dfrac{1-\ln x}{x^2}$, this function is increasing on $(0,\mathrm e]$. Observe that $$\sqrt 5<\sqrt 7<2.7<\mathrm e,\enspace\text{so }\enspace  f(\sqrt 5)<f(\sqrt 7). $$
A: Consider the function $f(x)=x^{1/x}$. So find derivative and set it zero to get increasing interval. $$f' (x)= \frac{x^{1/x-1}}{x} -\frac{x^{1/x} \ln(x)}{x^2}$$
From here, note that only root of $f'(x)$ is $e$, and $f$ is increasing for $x<e$. Since $e>\sqrt 7 > \sqrt 5$ we can say
$$f(\sqrt7) > f(\sqrt5 ) \\
 \implies \sqrt{7}^{1/\sqrt 7} > \sqrt5 ^{1/\sqrt 5}$$
Now raise both sides to power $\sqrt 5 \sqrt 7$ to get the desired result as $$\sqrt 7 ^ \sqrt 5 > \sqrt 5 ^ \sqrt 7$$
A: write your inequality in the form
$$\frac{\sqrt{5}}{\ln(\sqrt{5})}>\frac{\sqrt{7}}{\ln(\sqrt{7})}$$ and consider $$f(x)=\frac{x}{\ln(\sqrt{x})}$$
A: $$7^5=16807>15625=5^6\\7^5>5^6\\\sqrt7^5=(\sqrt7^{\sqrt 5})^{\sqrt 5}>\sqrt5^6=\sqrt5^{\sqrt {36}}>\sqrt5^{\sqrt {35}}=(\sqrt5^{\sqrt {7}})^{\sqrt{5}}\\\sqrt7^{\sqrt 5}>\sqrt5^{\sqrt 7}$$
