Điều kiện x > 0
Ta có:
\(x=\sqrt{x-\dfrac{1}{x}}\sqrt{1-\dfrac{1}{x}}\)
\(\Leftrightarrow1=\dfrac{1}{\sqrt{x}}\left(1-\dfrac{1}{x^2}\right)+\dfrac{1}{x}\left(1-\dfrac{1}{x}\right)\)
Áp dụng bunhia ta có:
\(\dfrac{1}{\sqrt{x}}\left(1-\dfrac{1}{x^2}\right)+\dfrac{1}{x}\left(1-\dfrac{1}{x}\right)\le\sqrt{\left(\dfrac{1}{x}+1-\dfrac{1}{x}\right)\left(\dfrac{1}{x^2}+1-\dfrac{1}{x^2}\right)}=1\)
Dấu = xảy ra khi
\(\dfrac{1}{\sqrt{x}}.\dfrac{1}{x}=\sqrt{1-\dfrac{1}{x}}.\sqrt{1-\dfrac{1}{x^2}}\)
\(\Leftrightarrow x^3-x^2-x=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)
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