minimum value of $\displaystyle f(x) = \frac{(1+x)^{0.8}}{1+x^{0.8}}$ without derivative minimum value of $\displaystyle f(x) = \frac{(1+x)^{0.8}}{1+x^{0.8}},x\geq 0$ without derivative
Binomial expansion of $\displaystyle (1+x)^{0.8} = 1+0.8 x-\frac{(0.8\cdot (0.8-1)x^2)}{2}+\cdots $
but from above does not get anything
could some help me with this

Raise your function to the power of five:
$$(f(x))^5=\left(\frac{(1+x)^{0.8}}{1+x^{0.8}}\right)^5$$
$$(f(x))^5=\frac{(1+x)^4}{1+5x^{0.8}+10x^{1.6}+10x^{2.4}+5x^{3.2}+x^4}$$
$$(f(x))^5=\frac{1+4x+6x^2+4x^3+x^4}{1+5x^{0.8}+10x^{1.6}+10x^{2.4}+5x^{3.2}+x^4}$$
$$(f(x))^5=\frac{x^{-2}+4x^{-1}+6+4x+x^2}{x^{-2}+5x^{-1.2}+10x^{-0.4}+10x^{0.4}+5x^{1.2}+x^2}$$
As $(f(x))^5=(f(x^{-1}))^5$ then you will have a minimum (or maximum) when $x=x^{-1}$, i.e. when $x=1$ (ignoring negative values as its unclear which root you would want for $x<0$).
