Given are positive rational numbers `a ,b , c` such that `a+b+c=1,` then prove that `a^a b^bc^c+a^bb^cc^a+a^c b^a c^blt=1.`

Explanation

Using weighted A.M ., G.M. inequality, we have
` (ca+ab+bc)/(a+b+c) ge (a^cb^ac^b)^((1)/(a+b+c))`
or `ca+ab+bc gr (a^cb^ac^b)`
Similarly,
` (ba+cb+ac)/(b+c+a) ge (a^b b^c c^a)^((1)/(a+b+c)) or ba+cb+ac ge (a^b b^c c^a)`
`(aa+b b+c c)/(b+c+a)ge(a^ab^bc^c)^((1)/(a+b+c))or aa +b b +c c ge (a^ab^bc^c)`
Adding togather, we get
` (a+b+c)(a+b+c)ge a^a b^b c^c +a^b +b^c +c^a + a^c +b^a+c^b`
`rArr 1 ge a^a b^b c^c+a^b b^c c^a +a^c b^a c^b`