$$\eqalign{
& bai\;nay\;thi\;cung\;don\;gian\;thoi\;nhung\;phuong\;phap\;thi\;hoi\;la\; \cr
& \left( {phuong\;phap\;thi\;ban\;co\;the\;tham\;khao\;trong\;cuon\;sach\;sang\;tao\;bdt\;cua\;pham\;kim\;hung\;} \right) \cr
& giai\;nhu\;sau: \cr
& \left( {phan\;nay\;la\;lam\;ra\;nhap\;nhe} \right) \cr
& de\;thay\;dau\;bang\;xay\;ra\;khia = b = c = 1 \cr
& khi\;do\;{1 \over {{a^2} + 1}} = {1 \over 2} \cr
& \to can\;tim\;so\;n\;\left( {n\;la\;so\;thuc} \right)sao\;cho\; \cr
& {1 \over {{a^2} + 1}} \ge n\left( {a - 1} \right) + {1 \over 2} \cr
& \leftrightarrow {{2 - {a^2} - 1} \over {2\left( {{a^2} + 1} \right)}} \ge n\left( {a - 1} \right) \cr
& \leftrightarrow {{\left( {1 - a} \right)\left( {1 + a} \right)} \over {2\left( {{a^2} + 1} \right)}} \ge n\left( {a - 1} \right) \leftrightarrow \left( {1 - a} \right)\left( {{{1 + a} \over {2{a^2} + 2}} + n} \right) \ge 0 \cr
& khi\;a = 1 \to {{1 + a} \over {2{a^2} + 2}} + n = {1 \over 2} + n = 0 \to n = - {1 \over 2}\;\left( {day\;chinh\;la\;gia\;tri\;n\;can\;tim} \right) \cr
& \to bieu\;thuc\;sau\;se\;dung:{1 \over {{a^2} + 1}} \ge {1 \over 2} - {1 \over 2}\left( {a - 1} \right) = 1 - {a \over 2} \cr
& \to sau\;khi\;tim\;dc\;n\;thi\;chung\;minh\;nhu\;sau \cr
& ta\;co: \cr
& {1 \over {{a^2} + 1}} - \left( {1 - {a \over 2}} \right) \cr
& = {{2 + a\left( {{a^2} + 1} \right) - 2{a^2} - 2} \over {2{a^2} + 2}} = {{{a^3} - 2{a^2} + a} \over {2{a^2} + 2}} = {{a{{\left( {a - 1} \right)}^2}} \over {2{a^2} + 2}} \ge 0\;\forall a > 0\;dau\; = \leftrightarrow a = 1 \cr
& \to {1 \over {{a^2} + 1}} \ge 1 - {a \over 2} \cr
& tuong\;tu \cr
& {1 \over {{b^2} + 1}} \ge 1 - {b \over 2} \cr
& {1 \over {{c^2} + 1}} \ge 1 - {c \over 2} \cr
& \to VT \ge 3 - {{\left( {a + b + c} \right)} \over 2} = {3 \over 2} \cr
& dau = \leftrightarrow a = b = c = 1 \cr
& \cr
& HOAC\;DUNG\;KI\;THUAT\;COSI\;NGUOC\;DAU: \cr
& {1 \over {{a^2} + 1}} = 1 - {{{a^2}} \over {{a^2} + 1}} \ge 1 - {{{a^2}} \over {2a}}\;\left( {{a^2} + 1 \ge 2a \to {1 \over {2a}} \ge {1 \over {{a^2} + 1}} \to {{{a^2}} \over {2a}} \ge {{{a^2}} \over {{a^2} + 1}} \to - {{{a^2}} \over {2a}} \le - {{{a^2}} \over {{a^2} + 1}}} \right) = 1 - {a \over 2} \cr} $$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn