Give: \(a+b+c\ge abc\).Prove that:

\(a^2+b^2+c^2\ge abc\)

Solve these equations:

a) \(\dfrac{6}{\left(x+1\right)\left(x+2\right)}+\dfrac{8}{\left(x-1\right)\left(x+4\right)}=1\)

b)\(\dfrac{x^2-3x+5}{x^2-4x+5}+\dfrac{x^2-5x+5}{x^2-6x+5}=\dfrac{-1}{4}\)