Ta có: pt ⇔√(x−3√22)2+92+√(x−2√2)2+8=5Trong Oxy, xét 2 vecto →u=(x−3√22;3√22),→v(2√2−x;2√2)
Suy ra: {/→u/=......./→v/=.......⇒{→u+→v=...../→u+→v/=.......=5
Mà ta luôn có: /→u/+/→v/≥/→u+→v|
⇔........⇒VT≥VP.
→.............
Đẳng thức khi: x−3√222√2−x=34→............