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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

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    $\begingroup$ For clarification can you give the domain you are interested in working on? Given the tag you have selected I'm assuming you are looking at $x\ge0$ as if $x<0$ then you need to define which root you mean by $x^{0.8}$. $\endgroup$ – Ian Miller Dec 31 '16 at 13:16
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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 $x<0$).

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  • $\begingroup$ Ian Miller in second last line , have u used arithmetic geometric inequality $\endgroup$ – DXT Dec 31 '16 at 13:43
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    $\begingroup$ Second last line I just divided top and bottom by $x^2$. $\endgroup$ – Ian Miller Dec 31 '16 at 17:09
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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.

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    $\begingroup$ I might be slow on this last day of the year, but does it follow from the parity directly that $u=0$ is a minimum? $\endgroup$ – mickep Dec 31 '16 at 15:28
  • $\begingroup$ @mickep: since $\frac{d}{du}\frac{\cosh(u/2)^{4/5}}{\cosh(2u/5)}$ does not vanish anywhere else, yes. $\endgroup$ – Jack D'Aurizio Dec 31 '16 at 15:29
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    $\begingroup$ Indeed. But the problem asked to solve this without derivatives. $\endgroup$ – mickep Dec 31 '16 at 15:32
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    $\begingroup$ @mickep: then it is enough to consider that by convexity the ratio between $\cosh(u/2)^{4/5}$ and $\cosh(2u/5)$ is increasing on $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Dec 31 '16 at 15:33
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    $\begingroup$ Hehe, well, OK, I buy that :) And cheers, I've enjoyed many of your answers this year. $\endgroup$ – mickep Dec 31 '16 at 15:36
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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.

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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 $x=1$.

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    $\begingroup$ cuteboy please explain $\displaystyle \left(\frac{1}{2}+\frac{x}{2}\right)^{0.8}\geq \left(\frac{1}{2}\cdot 1^{0.8}+\frac{1}{2}\cdot x^{0.8}\right)$ $\endgroup$ – DXT Dec 31 '16 at 13:30
  • $\begingroup$ @DURGESHTIWARI this is property of concave function. $\endgroup$ – ehochix Dec 31 '16 at 13:31
  • $\begingroup$ Perhaps the OP means why is it concave?.. $\endgroup$ – Ian Miller Dec 31 '16 at 13:31
  • $\begingroup$ @DURGESHTIWARI ok, a typo, the function $x^{0.8}$ is concave. $\endgroup$ – ehochix Dec 31 '16 at 13:31
  • $\begingroup$ @IanMiller ok, then it is concave since its 2nd derivative is negative on the domain. $\endgroup$ – ehochix Dec 31 '16 at 13:32

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