Đáp án:
Giải thích các bước giải:
$\dfrac{1}{a_{1}} - 1 = \dfrac{1 - a_{1}}{a_{1}} $
$ = \dfrac{a_{2} + a_{3} + a_{4} + a_{5}}{a_{1}} ≥ \dfrac{4\sqrt[4]{a_{2}a_{3}a_{4}a_{5}}}{a_{1}} (1)$
$\dfrac{1}{a_{2}} - 1 = \dfrac{1 - a_{2}}{a_{2}} $
$ = \dfrac{a_{1} + a_{3} + a_{4} + a_{5}}{a_{2}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{3}a_{4}a_{5}}}{a_{2}} (2)$
$\dfrac{1}{a_{3}} - 1 = \dfrac{1 - a_{3}}{a_{3}} $
$ = \dfrac{a_{1} + a_{2} + a_{4} + a_{5}}{a_{1}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{4}a_{5}}}{a_{3}} (3)$
$\dfrac{1}{a_{4}} - 1 = \dfrac{1 - a_{4}}{a_{4}} $
$ = \dfrac{a_{1} + a_{2} + a_{3} + a_{5}}{a_{4}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{3}a_{5}}}{a_{4}} (4)$
$\dfrac{1}{a_{5}} - 1 = \dfrac{1 - a_{5}}{a_{5}} $
$ = \dfrac{a_{1} + a_{2} + a_{3} + a_{4}}{a_{5}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{3}a_{4}}}{a_{5}} (5)$
$(1).(2).(3).(4).(5):$
$ (\dfrac{1}{a_{1}} - 1)(\dfrac{1}{a_{2}} - 1)(\dfrac{1}{a_{3}} - 1)(\dfrac{1}{a_{4}} - 1)(\dfrac{1}{a_{1}} - 1) ≥ 4^{5}.\dfrac{a_{1}a_{2}a_{3}a_{4}a_{5}}{a_{1}a_{2}a_{3}a_{4}a_{5}} = 1024$
