Cho $a \ge b > 0$. Khẳng định nào đúng?

$P = \;\dfrac{1}{a} + \dfrac{1}{b} - \dfrac{4}{{a + b}}$$= \dfrac{{a + b}}{{ab}} - \dfrac{4}{{a + b}}\\ = \dfrac{{{{\left( {a + b} \right)}^2} - 4ab}}{{ab\left( {a + b} \right)}}\;\; = \dfrac{{{a^2} + 2ab + {b^2} - 4ab}}{{ab\left( {a + b} \right)}}\;\\ = \dfrac{{{a^2} - 2ab + {b^2}}}{{ab\left( {a + b} \right)}} = \dfrac{{{{\left( {a - b} \right)}^2}}}{{ab\left( {a + b} \right)}}$

Do $a + b > 0;\;ab > 0$ và ${\left( {a - b} \right)^2} \ge 0\;\;\;\forall \;a,\;b$ nên $\dfrac{{{{\left( {a - b} \right)}^2}}}{{ab\left( {a + b} \right)}} \ge 0 \Rightarrow P \ge 0$ hay $\dfrac{1}{a} + \dfrac{1}{b} \ge \dfrac{4}{{a + b}}$.