Bài 13 trang 222 SGK Đại số 10 Nâng cao

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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