\(A=\dfrac{1}{bc}+\dfrac{1}{ac}\)
\(A\ge\dfrac{4}{\left(a+b\right)c}=\dfrac{4}{\left(1-c\right)c}\ge\dfrac{4}{\dfrac{1}{4}}=16\)
Dấu bằng xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}a=b=\dfrac{1}{4}\\c=\dfrac{1}{2}\end{matrix}\right.\)
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