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
 A: If you set $x=e^u$ you get:
$$ f(x) = \frac{(1+e^u)^{4/5}}{1+e^{4u/5}} = 2^{-1/5}\cdot\frac{\cosh(u/2)^{4/5}}{\cosh(2u/5)} $$
that is an even function of $u$. It follows that the minimum of $f$ is achieved at $u=0$, i.e. at $x=1$.
It is enough to show that $\log\cosh\frac{4x}{5}\leq \frac{4}{5}\log\cosh x$ for any $x\geq 0$. That follows from $\tanh\frac{4x}{5}\leq\tanh x$, that is trivial, through termwise integration.
A: 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 $x<0$).
A: By Power Means inequality $\left(\frac{1+x}{2}\right)^{\frac{4}{5}}\geq\frac{1+x^{\frac{4}{5}}}{2}$, which gives the answer.
A: Let's assume that $x\geq 0$, otherwise the root is not well defined. Now the function $x^{0.8}$ is concave, so $(1+x)^{0.8}=2^{0.8}(\frac{1}{2}+\frac{x}{2})^{0.8}\geq 2^{0.8}(\frac{1}{2}\times1^{0.8}+\frac{1}{2}\times x^{0.8})=2^{-0.2}(1+x^{0.8})$, so the minimum is $2^{-0.2}$ when $x=1$.
