Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$

Question

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.

Answer 1

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.

Answer 2

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 $a=b=\frac 13,\,c=3.$

Answer 3

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 $c$.