Show $ 5 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ≥ (1+a)(1 +b)(1+c)$ when $abc =1$ Just an inequality that I couldn't solve:
For $a,b,c$ positive reals such that $abc = 1$, show
$$5 + \frac{a}{b} +  \frac{b}{c}  + \frac{c}{a} \ge (1 + a)(1 + b)(1 + c).$$
WhatI have done so far: I have shown this is equivalent, after some algebra, to $3+a^2c+b^2a + c^2b \ge ab+bc+ac + a + b + c$, and I have absolutely no idea where to go from there.

Substitute $a = u/v, b = v/w, c = w/ u$ to get rid of the constraint $abc=1$, and you have the equivalent inequality
$$\sum_{cyc} u^3v^3 + 3u^2v^2w^2 \ge uvw \sum_{cyc} uv(u+v)$$
Now let $x = uv, y = vw, z = wu$ and we get in these new variables the inequality:
$$\sum_{cyc} x^3 + 3xyz \ge  \sum_{cyc} xy(x+y)$$
which is the familiar Schur's inequality.
