Which one is greater $\left(5/2\right)^{2/5}$ or $\left(7/2\right)^{2/7}$? I have a question here: 
Which of $\left(5/2\right)^{2/5}$ and $\left(7/2\right)^{2/7}$ is greater?
I tried comparison by the function $y=x^{1/x}$ and found the derivative as follows 
$$\frac{\partial y}{\partial x}=\frac{1}{x}(x^{\frac{1}{x}-1})+x^{1/x}\ln x=x^{1/x}(1/x^2+\ln x)$$
I got stuck here, what to do next?
My teacher told me the answer is $\left(5/2\right)^{2/5}$. I want to know how it is. 

It is enough to notice that:
$$ f(x) = \log\left(x^{\frac{1}{x}}\right) = \frac{\log x}{x} \tag{1}$$
fulfills:
$$ f'(x) = \frac{1-\log x}{x^2},\qquad f''(x) = \frac{2\log x-3}{x^3} \tag{2}$$
hence $f(x)$ is a concave function over the interval $I=\left[\frac{5}{2},\frac{7}{2}\right]$, increasing from $\frac{5}{2}$ to $e$ and decreasing from $e$ to $\frac{7}{2}$. Since the value of $f'''(x)=\frac{11-6\log x}{x^4}$ on $I$ is positive and quite small, the graph of $f(x)$ over $I$ is essentially the graph of a concave-down parabola with vertex at $x=e$. Since $\frac{5}{2}$ is the endpoint of $I$ closer to $e$,

$$ f\left(\frac{5}{2}\right) > f\left(\frac{7}{2}\right).\tag{3}$$

To prove it in a more formal way, compute the second-order Taylor expansion of $f(x)$ centered at $x=e$ and use Lagrange's remainder:
$$f(x) = f(e)+f'(e)(x-e)+\frac{f''(e)}{2}(x-e)^2+\frac{f'''(\xi)}{6}(x-e)^3,\qquad \xi\in I.\tag{4}$$
That gives $f(5/2)>\frac{73}{200}>f(7/2)$ and we are done.

An alternative to prove $5^7>4\cdot 7^5$ (that is equivalent to the previous claim) is to prove:
$$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6<\frac{35}{4}\tag{5}$$
by exploiting the Taylor series of the function $g(x)=\left(\frac{1+x}{1-x}\right)^6$:
$$ g(x) = 1+\sum_{n\geq 1}\frac{4n}{15}(23+20n^2+2n^4)\,x^n \tag{6} $$
and proving the inequality:
$$ \forall x\in(0,1),\qquad \left(\frac{1+\frac{x}{6}}{1-\frac{x}{6}}\right)^6\leq 1+2x+2x^2+3x^3\tag{7}$$
from which:
$$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6\leq 8.\tag{8}$$
