An idea to find range with $|\frac{a}{b}+\frac{b}{a}|\geq2$

Question

How can find range of $f(x)=\frac {4x^2+8x+1}x$ by $$|\frac{a}{b}+\frac{b}{a}|\geq2$$ I can find range of that function with inverse function ,but how can I apply this inequality to the function ? thanks for you idea(s).

Answer 1

First simplify $f(x)=\frac {4x^2+8x+1}x=4x+8+\frac 1x$ so $$f(x)-8=4x+\frac {1}{x} \to $$multiply by $\frac 12$ $$\frac{1}{2}(f(x)-8)=2x+\frac{1}{2x}=\frac{a}{b}+\frac{b}{a}\\ |\frac{1}{2}(f(x)-8)|\geq 2\\|f(x)-8|\geq 4$$