Proving $f(2) \geq 27$ with Positive Coefficients

[Albert1]

given :$f(x)=x^3+ax^2+bx+1$ with $a\geq 0 \,\, and \,\, b\geq 0$ and also all of the three roots of $f(x)=0$ are real numbers,

prove $f(2)\geq 27$

 

[Albert1]

given :$f(x)=x^3+ax^2+bx+1$ with $a\geq 0 \,\, and \,\, b\geq 0$ and also all of the three roots of $f(x)=0$ are real numbers,

prove $f(2)\geq 27$

Spoiler

using Descartes' Rule of Signs all of the three roots of $f(x)=0---(1)$ must be negative
and let $-x_1,-x_2,-x_3$ be the solutions of $(1)$
(here $x_1,x_2,x_3>0$)
from Vieta's formulas and $AP\geq GP$ we have:
$x_1x_2x_3=1---(2)$
from $(2)$
$a=x_1+x_2+x_3\geq 3\sqrt[3]{x_1x_2x_3}=3---(3)\\
b=x_1x_2+x_1x_3+x_2x_3\geq 3\sqrt[3]{x_1^2x_2^2x_3^2}=3---(4)\\$
from $(3)(4):f(2)\geq 27=2^3+3\times 2^2+3\times 2+1=8+12+6+1$