Bài 13 trang 222 SGK Đại số 10 Nâng cao
Chứng minh rằng: ...
Chứng minh rằng:
Chứng minh rằng:
a) ({{{a^2} + 6} over {sqrt {{a^2} + 2} }} ge 4,,,,(a in R))
b) ({{{a^2}} over {{b^2}}} + {{{b^2}} over {{c^2}}} + {{{c^2}} over {{a^2}}} ge {a over c} + {c over b} + {b over a},,,(a,,b,,c, in R))
Đáp án
a) Áp dụng bất đẳng thức Cô-si, ta có:
({{{a^2} + 6} over {sqrt {{a^2} + 2} }} = {{({a^2} + 2) + 4} over {sqrt {{a^2} + 2} }} = sqrt {{a^2} + 2} + {4 over {sqrt {{a^2} + 2} }} ge )
(2sqrt {sqrt {{a^2} + 2} .{4 over {sqrt {{a^2} + 2} }}} = 4)
b) Ta có:
({{{a^2}} over {{b^2}}} + {{{b^2}} over {{c^2}}} ge 2sqrt {{{{a^2}} over {{b^2}}}.{{{b^2}} over {{c^2}}}} = 2|{a over c}|, ge {{2a} over c})
Tương tự ta có:
(left{ matrix{
{{{b^2}} over {{c^2}}} + {{{c^2}} over {{a^2}}} ge 2{b over a} hfill cr
{{{c^2}} over {{a^2}}} + {{{a^2}} over {{b^2}}} ge 2{c over b} hfill cr}
ight.)
Từ đó suy ra: (2({{{a^2}} over {{b^2}}} + {{{b^2}} over {{c^2}}} + {{{c^2}} over {{a^2}}}) ge 2({a over c} + {c over b} + {b over a}))
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