Question

Answer 1

P = \(\dfrac{1}{a^2-ab+b^2}+\dfrac{3}{ab}=\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{ab}+\dfrac{1}{ab}+\dfrac{1}{ab}\)

\(\ge\dfrac{\left(1+1+1+1\right)^2}{a^2-ab+b^2+ab+ab+ab}=\dfrac{16}{\left(a+b\right)^2}=\dfrac{16}{2^2}=4\)

"=" khi a = b = 1