- #1

- 2

- 0

Pls help me to prove the following inequality:

\begin{align*}

3 \left(\dfrac{a}{b} + \dfrac{b}{a} + \dfrac{b}{c} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{c}{a}\right) &+ \left( 1 + a\right) \left( 1+b \right)\left( 1+c \right)\left(\dfrac{c}{b}+\dfrac{c}{a} \right) \left( \dfrac{b}{a}+\dfrac{b}{c} \right) \left( \dfrac{a}{b}+\dfrac{a}{c} \right) \\ & \geq 6abc+6+9(ab+bc+ac+a+b+c)+3\left( \dfrac{ab}{c} +\dfrac{bc}{a}+\dfrac{ac}{b}\right)

\end{align*}

with ##a, b, c## are positive reals

If it helps, I know the equality occurs when ##a=b=c=2## (although I'm not sure if it's the only one).

Also, can any one helps to prove ##(2, 2, 2)## is the only point at which equality occurs.

Thanks a lot...

Any hint is appreciated...

\begin{align*}

3 \left(\dfrac{a}{b} + \dfrac{b}{a} + \dfrac{b}{c} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{c}{a}\right) &+ \left( 1 + a\right) \left( 1+b \right)\left( 1+c \right)\left(\dfrac{c}{b}+\dfrac{c}{a} \right) \left( \dfrac{b}{a}+\dfrac{b}{c} \right) \left( \dfrac{a}{b}+\dfrac{a}{c} \right) \\ & \geq 6abc+6+9(ab+bc+ac+a+b+c)+3\left( \dfrac{ab}{c} +\dfrac{bc}{a}+\dfrac{ac}{b}\right)

\end{align*}

with ##a, b, c## are positive reals

If it helps, I know the equality occurs when ##a=b=c=2## (although I'm not sure if it's the only one).

Also, can any one helps to prove ##(2, 2, 2)## is the only point at which equality occurs.

Thanks a lot...

Any hint is appreciated...

Last edited by a moderator: