Is it possible to prove that $$t^\frac{1}{t}>y^\frac{1}{y}$$ knowing that $e\leq{t} <y$ and without looking at the function $$x^\frac{1}{x}$$?
I already tried with exponentials, logs, properties of logs, but nothing seems to work. In fact, the above writing is another form for saying $t^y>y^t$, which seems impossible to solve with standard methods. Are there any suggestions?
You can show that it is true for $t \ge e$, but you have to look at the function or its log.
As Vanwij suggested, let $f(x) = x^{1/x}$, though it is easier with its log $g(x) = \dfrac{\ln(x)}{x} $.
Then $g'(x) =\dfrac{1-\ln(x)}{x^2} $ so $g'(x) = 0$ for $x = e$ and $g'(x) < 0$ for $x > e$.
Therefore if $e \le t \lt y$ then $g(t) > g(y) $.
However, you can show that $x^{1/x}$ has its max just from $e^x \gt 1+x$ except at $x = 0$.
(This is not original.)
If $x \ne e$ we have $e^{\frac{x-e}{e}} \gt 1+\frac{x-e}{e} =1+\frac{x}{e}-1 =\frac{x}{e} $ so $e^{\frac{x}{e}-1} \gt \frac{x}{e} $ or $e^{\frac{x}{e}} \gt x $ or $e^{1/e} > x^{1/x}$.