Since $\sum_{i=1}^6 a_i^3=\sum_{i=1}^6 b_i^3=\sum_{i=1}^6 c_i^3=\sum_{k=1}^6 k^3=\left\{\frac{6(6+1)}{2}\right\}^2=21^2$, $\left(\sum_{i=1}^6 a_i b_i c_i\right)^3\leq \left(\sum_{i=1}^6 a_i^3\right)\left(\sum_{i=1}^6 b_i^3\right)\left(\sum_{i=1}^6 c_i^3\right)$, so $\sum_{i=1}^6 a_i b_i c_i\leq 21^2=441.$ Are you sure that minimum value, zhaoli?

It's a TST; I expect that they give harder problems than a trivial application of Holder. Even the rearrangement inequality gives the same result.

I think the answer is 162: 3 groups of (6*5*1=)30 and 3 groups of (4*3*2=)24. Is this right zhaoli?

al.M.V. wrote: I think the answer is 162: 3 groups of (6*5*1=)30 and 3 groups of (4*3*2=)24. Is this right zhaoli? Yes,you are right.

From the conditions of this problem, we know that (1) $a_i$, $i=1, 2, 3, 4, 5, 6$, are all positive integers, and pairwise distinct. So does $b_i$, $i=1, 2, 3, 4, 5, 6$; and $c_i$, $i=1, 2, 3, 4, 5, 6$. (2) They are finite permutations of $1, 2, 3, 4, 5, 6$, there must exist minimum value.

can any one give the complete solution?

yeah, can anyone?

zhaoli wrote: Let $a_{1}$, $a_{2}$, …, $a_{6}$; $b_{1}$, $b_{2}$, …, $b_{6}$ and $c_{1}$, $c_{2}$, …, $c_{6}$ are all permutations of $1$, $2$, …, $6$, respectively. Find the minimum value of $\sum_{i=1}^{6}a_{i}b_{i}c_{i}$. By AM-GM inequality $\sum_{i=1}^{6}a_{i}b_{i}c_{i}\ge 6\sqrt{6!}\Leftrightarrow \sum_{i=1}^{6}a_{i}b_{i}c_{i}\ge 161$ Now the arrangement $(a_1,b_1,c_1);(a_2,b_2,c_2);...;(a_6,b_6,c_6)=(6,4,1);(4,1,6);(1,6,4);(5,3,2);(3,2,5);(2,5,3)$ gives $\sum_{i=1}^{6}a_{i}b_{i}c_{i}=162$ Now suppose that $\sum_{i=1}^{6}a_{i}b_{i}c_{i}=161$ then at least one of $a_ib_ic_i$ is odd, Suppose that it is $a_1b_1c_1$ then $161\ge a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}$ From here $a_1b_1c_1> 11$ so the only possibility is $a_1b_1c_1=5\cdot 3\cdot 1=15$ which gives $a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}>161$ hence $min\{\sum_{i=1}^{6}a_{i}b_{i}c_{i}\}=162$

hardsoul wrote: so the only possibility is $a_1b_1c_1=5\cdot 3\cdot 1=15$ Why is that, hardsoul? I fail to see why could you explain better?

perfect_radio wrote: hardsoul wrote: so the only possibility is $a_1b_1c_1=5\cdot 3\cdot 1=15$ Why is that, hardsoul? I fail to see why could you explain better? You are right we must also check the cases $a_1b_1c_1=5\cdot 3\cdot 1=15,$ $a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}=165.71...>161$ $a_1b_1c_1=5\cdot 5\cdot 1=25,$ $a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}=161.07...>161$ $a_1b_1c_1=5\cdot 3\cdot 3=45,$ $a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}=165,98...>161$ $a_1b_1c_1=5\cdot 5\cdot 5=125,$ $a_1b_1c_1+5\left(\frac{(6!)^3}{a_1b_1c_1}\right)^{1/5}=223,62...>161$

actually, i think there are more posibilities (like 3,3,3). only a few are under 12. anyway, this problem is much too "computational". i was hoping you had an argument which avoids all that

perfect_radio wrote: actually, i think there are more posibilities (like 3,3,3). .. It was the last one .Although I don't know whether there's another way to avoid comput.,I think that to check if these values sat. the ineq. is not so difficult...

hardsoul wrote: Now suppose that $\sum_{i=1}^{6}a_{i}b_{i}c_{i}=161$ This is a flaw in your proof. There could be an arrangement of $a_i b_i c_i$ with sum less than $162$, and none of sum $161$. --- Firstly we remark that the product of all terms of the sum is equal to $6!^3$. Then we see that the numbers that can be written as a product of three positive integers at most $7$ are \[ 1,2,\cdots,20,24,25,27,30,32,36,\cdots,216. \] If one of $a_i b_i c_i$ is less than 24, then it is at most $20$, thus \[ \sum \left(a_i b_i c_i \right) \geq 20 + 5 \sqrt[5]{\frac{6!^3}{20}} > 20 + 5*28,4 = 162. \] If one is at least $36$ then \[ \sum \left(a_i b_i c_i \right) \geq 36 + 5 \sqrt[5]{\frac{6!^3}{36}} > 36 + 5*26 > 162. \] Thus we just have to study the case where all the $a_i b_i c_i$ are in $E=\{24,25,27,30,32\}$. We want to write $6!^3$ as a product of six numbers of $E$. These numbers will have to contain three times the factor $5$, so either we write $6!=30*30*30*x*y*z$ or $6!^3=25*30*w*x*y*z$. In the first case, to minimize the sum $30+30+30+x+y+z$ we take $x=y=z=24$, which yields $30+30+30+24+24+24=162$. In the second case $wxyz=2^11*3^5$, and $w$,$x$,$y$ and $z$ are in $\{24,27,32\}$. We can easily investigate this case, which gives $24*24*27*32=2^11*3^5$, thus finally $24*24*25*27*30*32=6!^3$ and $24+24+25+27+30+32=162$. The minimal value of the sum is thus $162$, with a lot of equality cases.

toumaf wrote: hardsoul wrote: Now suppose that $\sum_{i=1}^{6}a_{i}b_{i}c_{i}=161$ This is a flaw in your proof. There could be an arrangement of $a_i b_i c_i$ with sum less than $162$, and none of sum $161$. --- Firstly we remark that the product of all terms of the sum is equal to $6!^3$. Then we see that the numbers that can be written as a product of three positive integers at most $7$ are \[ 1,2,\cdots,20,24,25,27,30,32,36,\cdots,216. \] If one of $a_i b_i c_i$ is less than 24, then it is at most $20$, thus \[ \sum \left(a_i b_i c_i \right) \geq 20 + 5 \sqrt[5]{\frac{6!^3}{20}} > 20 + 5*28,4 = 162. \] If one is at least $36$ then \[ \sum \left(a_i b_i c_i \right) \geq 36 + 5 \sqrt[5]{\frac{6!^3}{36}} > 36 + 5*26 > 162. \] Thus we just have to study the case where all the $a_i b_i c_i$ are in $E=\{24,25,27,30,32\}$. We want to write $6!^3$ as a product of six numbers of $E$. These numbers will have to contain three times the factor $5$, so either we write $6!=30*30*30*x*y*z$ or $6!^3=25*30*w*x*y*z$. In the first case, to minimize the sum $30+30+30+x+y+z$ we take $x=y=z=24$, which yields $30+30+30+24+24+24=162$. In the second case $wxyz=2^11*3^5$, and $w$,$x$,$y$ and $z$ are in $\{24,27,32\}$. We can easily investigate this case, which gives $24*24*27*32=2^11*3^5$, thus finally $24*24*25*27*30*32=6!^3$ and $24+24+25+27+30+32=162$. The minimal value of the sum is thus $162$, with a lot of equality cases. Thanks,I see.

Interesting problem!