IIT JEE , AIEEE Forum - Mathematics , Algebra

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edit: ans is 1 only..
do AM- GM
$\frac {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b}{3} \ge (a^{a+b+c}b^{a+b+c}c^{a+b+c})^{\frac 1 3}$
RHS is nothing but $(abc)^{\frac 1 3}$
but max value of abc is when a=b=c in the expression a+b+c = 1
so, max value of abc = 1/27
so max value of RHS = $(\frac 1 {27})^{\frac 1 3}$
$= \frac 1 3$

so $\frac {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b}{3} \ge = \frac 1 3$
$=> {a^ab^bc^c + a^bb^cc^a + a^cb^ac^b} \ge 1$
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