+) Ta có: [imath]a+b\geq 2\sqrt{ab}[/imath] ( cauchuy )
mà [imath]0<a+b\leq1[/imath]
[imath]\Rightarrow 1\geq 2\sqrt{ab}\Rightarrow \dfrac{1}{\sqrt{ab}}\geq 2[/imath]
+) Lại có:
[imath]a+\dfrac{1}{a}+b+\dfrac{1}{b}=a+\dfrac{1}{4a}+b+\dfrac{1}{4b}+\dfrac{3}{4}(\dfrac{1}{a}+\dfrac{1}{b})[/imath]
[imath]\geq 2\sqrt{a.\dfrac{1}{4a}}+2\sqrt{b.\dfrac{1}{4b}}+\dfrac{3}{4}.2\sqrt{\dfrac{1}{ab}}[/imath]
[imath]=2.\dfrac{1}{2}+2.\dfrac{1}{2}+\dfrac{3}{2}.\dfrac{1}{\sqrt{ab}}[/imath]
[imath]\geq 2+\dfrac{3}{2}.2=5[/imath] (dcpm)
+) Dấu [imath]"="[/imath] xảy ra khi [imath]a=b=\dfrac{1}{2}[/imath]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
