[toán 9] GTNN

D

demon311

Ta có:

$2\sqrt{ab} \le a+b \le 1 \\
\Rightarrow ab \le \dfrac{ 1}{4} \\
\Rightarrow a^2b^2 \le \dfrac{ 1}{16} \\
\Rightarrow 4ab \le \dfrac{ 1}{4ab}$

$S = \dfrac{ 1}{a^2+b^2}+\dfrac{ 1}{2ab}+\dfrac{ 1}{4ab}+4ab+\dfrac{ 1}{4ab} \\
\ge \dfrac{ 4}{(a+b)^2}+2\sqrt{ \dfrac{ 4ab}{4ab}}+\dfrac{ 1}{4\dfrac{ 1}{4}} \ge 4+2+1=7$
 
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