JOMPC-II

Wrong method :

Using AM-GM:

$\large{a+2015b\ge 2\sqrt { 2015ab } \\ b+2015c\ge 2\sqrt { 2015bc } \\ c+2015a\ge 2\sqrt { 2015ca } \\ \longrightarrow \left( a+2015b \right) \left( b+2015c \right) \left( c+2015a \right) \ge 8\sqrt { { 2015 }^{ 3 } } }$ .

But this is not possible because for equality to hold we must have $a=2015b, b=2015c, c=2015a$ which is absurd as it will give on multiplying $abc=(2015)^3 abc$

Correct method :

Using AM-GM:

$\large{a+\underbrace { b+b+......+b }_{ 2015\quad times } \ge 2016\sqrt [ 2016 ]{ a{ b }^{ 2015 } } \\ b+\underbrace { c+c+......+c }_{ 2015\quad times } \ge 2016\sqrt [ 2016 ]{ b{ c }^{ 2015 } } \\ c+\underbrace { a+a+......+a }_{ 2015\quad times } \ge 2016\sqrt [ 2016 ]{ c{ a }^{ 2015 } } \\ \longrightarrow \left( a+2015b \right) \left( b+2015c \right) \left( c+2015a \right) \ge { 2016 }^{ 3 }}$

And equality holds when $a=b=c=1$ .

My Method: :)

WLOG, let $a \geq b \geq c$ . By the extreme power of common sense, $2015 a \geq 2015b \geq 2015c$ .

Then, by Reverse-rearrangement inequality,

$(a+2015b)(b+2015c)(c+2015a) \geq (a+2015a)(b+2015b)(c+2015c)$

$(a+2015b)(b+2015c)(c+2015a) \geq (2016a)(2016b)(2016c)$

$(a+2015b)(b+2015c)(c+2015a) \geq 2016^3abc$ . Since abc = 1,

$(a+2015b)(b+2015c)(c+2015a) \geq 2016^3$ .

Then, finding the last three digits of $2016^3$ is the same as finding the last three digits of $16^3$ . So, last three digits of $16^3$ are the last three digits of $4096$ which are $\boxed{96}$

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By coincidence, I was studying the Reverse-rearrangement inequality when I saw this problem.