<s>[INST] Inequality with Algebra $\sup_{x,y,z\in A}\left(\left(13x+\frac{5}{x}\right)+\left(13y+\frac{5}{y}\right)+\left(13z+\frac{5}{z}\right)\right)= 63$ let $A=\{(x,y,z)|x,y,z\in [\frac{1}{2},2],xyz=1\}$
Maybe have
$$f=\sup_{x,y,z\in A}\left(\left(13x+\dfrac{5}{x}\right)+\left(13y+\dfrac{5}{y}\right)+\left(13z+\dfrac{5}{z}\right)\right)=  63?$$
Here the maximum of $f$ is $63$,reached at the point $(x,y,z)$ on the space surface $xyz=1$,and  $x=2,y=\dfrac{1}{2},z=1$
I attempted a proof by AM-GM inequality,But I am not able to solve the upper part,   this inequality when $x=2,y=\dfrac{1}{2},z=1$,then $f=63$
This problem  can use from space Analysis geometry? ,take $y=\dfrac{15}{3x}$
 [/INST] $p=x+y,xy=q, \dfrac{1}{2} \le z \le 2 \implies \dfrac{1}{2} \le q \le 2 $
$f=13p+\dfrac{5p+13}{q}+5q$
consider $\dfrac{A}{q}+Bq$ will get max at two ends, so $f(p,q)$ will get max at $q=2$ or $q= \dfrac{1}{2}$
when $q=2 ,y=\dfrac{2}{x}$, put it in $f$,it become $f(x)= \dfrac{A}{x}+Bx+C$,again, $f(x)$ will get max at two ends,so you will have $x=2$ or $x=\dfrac{1}{2}$ to check which one is bigger. 
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