Áp dụng BĐT BSC:
\(A=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\)
\(=\dfrac{\dfrac{1}{16}}{x}+\dfrac{\dfrac{1}{4}}{y}+\dfrac{1}{z}\)
\(\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x+y+z}=\dfrac{49}{16}\)
\(minA=\dfrac{49}{16}\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{1}{4}}{x}=\dfrac{\dfrac{1}{2}}{y}=\dfrac{1}{z}\\x+y+z=1\end{matrix}\right.\)
\(\Leftrightarrow\left(x;y;z\right)=\left(\dfrac{1}{7};\dfrac{2}{7};\dfrac{4}{7}\right)\)
\(P=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}+\dfrac{49}{16}-\dfrac{49}{16}\)
\(P=\left(\dfrac{1}{16x}+\dfrac{49x}{16}\right)+\left(\dfrac{1}{4y}+\dfrac{49y}{16}\right)+\left(\dfrac{1}{z}+\dfrac{49z}{16}\right)-\dfrac{49}{16}\)
\(P\ge2\sqrt{\dfrac{49x}{16x.16}}+2\sqrt{\dfrac{49y}{4y.16}}+2\sqrt{\dfrac{49z}{z.16}}-\dfrac{49}{16}=\dfrac{49}{16}\)
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