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:
\({{{a^2} + 6} \over {\sqrt {{a^2} + 2} }} = {{({a^2} + 2) + 4} \over {\sqrt {{a^2} + 2} }} = \sqrt {{a^2} + 2} + {4 \over {\sqrt {{a^2} + 2} }} \ge \)
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\(2\sqrt {\sqrt {{a^2} + 2} .{4 \over {\sqrt {{a^2} + 2} }}} = 4\)
b) Ta có:
\({{{a^2}} \over {{b^2}}} + {{{b^2}} \over {{c^2}}} \ge 2\sqrt {{{{a^2}} \over {{b^2}}}.{{{b^2}} \over {{c^2}}}} = 2|{a \over c}|\, \ge {{2a} \over c}\)
Tương tự ta có:
\(\left\{ \matrix{
{{{b^2}} \over {{c^2}}} + {{{c^2}} \over {{a^2}}} \ge 2{b \over a} \hfill \cr
{{{c^2}} \over {{a^2}}} + {{{a^2}} \over {{b^2}}} \ge 2{c \over b} \hfill \cr} \right.\)
Từ đó suy ra: \(2({{{a^2}} \over {{b^2}}} + {{{b^2}} \over {{c^2}}} + {{{c^2}} \over {{a^2}}}) \ge 2({a \over c} + {c \over b} + {b \over a})\)