Toán 9-bđt

E

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$A={a^2}+\dfrac{1}{16{a^2}}+{b^2}+\dfrac{1}{16{b^2}}+\dfrac{15}{16}(\dfrac{1}{{a^2}}+\dfrac{1}{{b^2}}) \ge 1+\dfrac{15}{16}.\dfrac{2}{ab} \ge 1+ \dfrac{15}{16}.\dfrac{2}{\dfrac{1}{4}}=8,5$

Dấu "=" xảy ra \Leftrightarrow $a=b=\dfrac{1}{2}$
 
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