Question
If a,b,c \in R^+, prove a^ab^bc^c\geq (abc)\tiny^{\dfrac{a+b+c}{3} }
If a,b,c \in R^+, prove a^ab^bc^c\geq (abc)\tiny^{\dfrac{a+b+c}{3} }
Let a\geq b\geq c
\dfrac{a^ab^bc^c}{ (abc)\tiny^{\dfrac{a+b+c}{3} } } = \dfrac{a^{3a}b^{3b}c^{3c}}{(abc)^{a+b+c}}
=\dfrac{a^{2a}b^{2b}c^{2c}}{a^{b+c}b^{a+c}c^{a+b}}
= \Big( \dfrac{a}{b} \Big) ^{a-b}\Big( \dfrac{b}{c} \Big) ^{b-c}\Big( \dfrac{a}{c} \Big) ^{a-c}
\geq 1