Ta có:
\(A=\dfrac{1}{2a+b}+\dfrac{1}{a+2b}\)
\(=\dfrac{1}{2a+16-a}+\dfrac{1}{16-b+2b}\)
\(=\dfrac{1}{a+16}+\dfrac{1}{b+16}\)
\(=\dfrac{a+b+32}{ab+16\left(a+b\right)+256}\)
\(=\dfrac{16+32}{ab+256+256}\)
\(=\dfrac{48}{ab+512}\)
\(\ge\dfrac{48}{\dfrac{\left(a+b\right)^2}{4}+512}\) (Cô - si)
\(=\dfrac{48}{\dfrac{256}{4}+512}\)
\(=\dfrac{1}{12}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=8\)
Vậy Min A = \(\dfrac{1}{12}\) \(\Leftrightarrow a=b=8\)
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