# Difficult inequality

LanNguyen
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...

Last edited by a moderator:

## Answers and Replies

Mentor
2021 Award
With ##f(a,b,c) = a^2b+a^2c+ab^2+ac^2+b^2c+bc^2## and ##g(a,b,c)=a^2b^2+a^2c^2+b^2c^2## we can write the inequality as
$$3 (f(a,b,c)-g(a,b,c)) +3a^2b^2c^2+3abc \geq (a+1)(b+1)(c+1)(7g(a,b,c)-f(a,b,c))$$
With ##P_0=abc\, , \,P_1= (a+1)(b+1)(c+1)## and ##P_2=(a+b)(a+c)(b+c)## we get
$$P_2(3+P_1) \geq 3g(a,b,c) -3P_0^2+3P_0+9P_0P_1$$
and if we write ##g(a,b,c) =3P_0Q_3\, , \,P_2=3P_0Q_2\, , \,P_1=Q_1## and ##Q_0=P_0=abc## we have
$$Q_2Q_1\geq -2Q_2+3Q_1+(Q_3-Q_0+1)$$
with symmetric functions ##Q_i##. The inequality now looks like a conic section. Maybe normalizing it gets a geometric intuition.

Another idea is to write the entire inequality in terms of the elementary symmetric functions:
##a+b+c\, , \,ab+bc+ac\, , \,abc## which are the coefficients in ##p(x)=(x-a)(x-b)(x-c)##. The statement then is about the zeroes of this polynomial.