Question

If a,b,c \in R^+, prove a^ab^bc^c\geq (abc)\tiny^{\dfrac{a+b+c}{3} }

Collected in the board: Inequality

Steven Zheng posted 1 month ago


Answer

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

Steven Zheng posted 1 month ago

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