MHB Determine the minimum value of an expression

[anemone]

If $x,\,y,\,z>0$ and $x+y+z=1$, find the minimum of $\left(x+\dfrac{1}{x}\right)^{10}+\left(y+\dfrac{1}{y}\right)^{10}+\left(z+\dfrac{1}{z}\right)^{10}$.

 

[Albert1]

If $x,\,y,\,z>0$ and $x+y+z=1$, find the minimum of $\left(x+\dfrac{1}{x}\right)^{10}+\left(y+\dfrac{1}{y}\right)^{10}+\left(z+\dfrac{1}{z}\right)^{10}$.

Spoiler

for $x,y,z>0$
by using AM-GM inequality,
the minimum value will hold when :$x+\dfrac {1}{x}=y+\dfrac {1}{y}=z+\dfrac {1}{z}$
or $x=y=z=\dfrac{1}{3}$
the minimum value =$3\times (\dfrac {10}{3})^{10}$

 

[anemone]

Thanks Albert for participating and your solution! :)

Here is another solution that I want to share:

Spoiler

Given that $0<x,\,y,\,z<1$.

If we let $f(x)=\left(x+\dfrac{1}{x}\right)^{10}$ on $I=90,\,1)$, then $f$ is strictly convex on I because $f''(x)=90\left(x+\dfrac{1}{x}\right)^{8}\left(1-\dfrac{1}{x^2}\right)^{2}+10\left(x+\dfrac{1}{x}\right)^{9}\left(\dfrac{2}{x^3}\right)>0$ for $x\in I$.

By Jensen's inequality,

$\begin{align*}3f\left(\dfrac{x+y+z}{3}\right)&=3f\left(\dfrac{1}{3}\right)\\&=\dfrac{10^{10}}{3^9}\\&\le f(x)+f(y)+f(z)\\&=\left(x+\dfrac{1}{x}\right)^{10}+\left(y+\dfrac{1}{y}\right)^{10}+\left(z+\dfrac{1}{z}\right)^{10}\end{align*}$

Therefore, the minimum is $\dfrac{10^{10}}{3^9}$, attained when $x=y=z=\dfrac{1}{3}$.