Prove Inequality: 3(ab+bc+ac)/abc + 6abc + 15 ≥ 27

[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 anyone helps to prove $(2, 2, 2)$ is the only point at which equality occurs.

Thanks a lot...

Any hint is appreciated...

 

[fresh_42]

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.