Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$
I wish to find a proof with $a:\neq {\rm mid}\left \{ a, b, c \right \},$ or another proof$?$
Actually$,$ I also found a proof true for all $a,b,c \in \Big[\dfrac{1}{3},3\Big],$ but very ugly.
After clearing the denominators$,$ need to prove$:$
$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$
but we have$:$
$$f=\dfrac{1}{32} \left( 3-a \right)  \left( 3-b \right)  \Big( c-\dfrac{1}{3} \Big) +
 \left( 3-a \right)  \left( a-\dfrac{1}{3} \right)  \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right)  \left( b-\dfrac{1}{3} \right)  \left( 
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right)  \left( 3-c
 \right)  \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right)  \left( 3-c
 \right)  \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right)  \left( c-\dfrac{1}{3}
 \right)  \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right) 
 \left( b-\dfrac{1}{3} \right)  \left( c-\dfrac{1}{3} \right) + \left( 3-b \right) 
 \left( b-\dfrac{1}{3} \right)  \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
 \left( 3-b \right)  \left( a-\dfrac{1}{3} \right)  \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
 \left( 3-a \right)  \left( c-\dfrac{1}{3} \right)  \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
 \right)  \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
 \left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right)  \left( 
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right)  \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right)  \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right)  \left( c-\dfrac{1}{3} \right) ^{2}$$
So we are done.
If you want to check my decomposition$,$ please see the text here.
 A: By AM-GM we have
$$
\frac{(a+b+c) + (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}.
$$
Note that by the assumption, we have
$$
3 + \frac{1}{3} \geq a + \frac{1}{a}
$$
and similarly for the other variables. Therefore
$$
3 \cdot \frac{10}{3} \cdot \frac{1}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)},
$$
as desired.
A: I found a better estimation
$$ (a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big)  \leqslant \frac{209}{9}.$$
Equality occur when $a=b=3,\,c=\frac 13$ or $a=b=\frac 13,\,c=3.$
A: Let $f(a,b,c)=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. Note that $f$ is concave of each variable (if other variables are fixed). Hence, since concave on $I$ fucntion attains its maximum at endpoint of $I$ (here $I=[m,M]=\left[\frac{1}{3},3\right]$)
$$
\max_{(a,b,c)\in I^3} f=\max_{(a,b,c)\in\{m,M\}^3} f.
$$
Thus, we need only to compute these 8 values and choose the maximal one.
Details: consider any point $(a,b,c)$, fix $b$ and $c$ and consider $f$ as a function of $a$. We obtain
$$
f(a,b,c)\leq\max\{f(m,b,c),f(M,b,c)\},
$$
so we can assume that $a\in\{m,M\}$. Now repeat this argument for $b$ and $c$.
