Giải thích các bước giải:
1.Ta có :
$L=\dfrac{x^2+6x+8}{x}=x+\dfrac{8}{x}+6\ge 2\sqrt{x.\dfrac{8}{x}}+6=4\sqrt{2}+6$
2.Ta có :
$P=\dfrac{x^2}{3}+\dfrac{x^2}{3}+\dfrac{x^2}{3}+\dfrac{1}{x^3}+\dfrac{1}{x^3}\ge 5\sqrt[5]{\dfrac{x^2}{3}.\dfrac{x^2}{3}.\dfrac{x^2}{3}.\dfrac{1}{x^3}.\dfrac{1}{x^3}}=5\sqrt[5]{\dfrac{1}{27}}$