Dễ thấy với \(x=2\) ta có VT > VP.
Bạn xem lại đề.
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\(3\left(x^2-\frac{1}{x^2}\right)< 2\left(x^3-\frac{1}{x^3}\right)\)
\(\Leftrightarrow3\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)-2\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)< 0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)\left[3\left(x+\frac{1}{x}\right)-2\left(x^2+\frac{1}{x^2}+1\right)\right]< 0\)
Do \(x>1\Leftrightarrow x^2>1\Leftrightarrow x^2-1>0\)
\(\Rightarrow x-\frac{1}{x}=\frac{x^2-1}{x}>0\forall x>1\)
\(pt\Leftrightarrow3\left(x+\frac{1}{x}\right)-2\left(x^2+\frac{1}{x^2}+1\right)< 0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)-2\left(x^2+2+\frac{1}{x^2}-1\right)< 0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)-2\left[\left(x+\frac{1}{x}\right)^2-1\right]< 0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)-2\left(x+\frac{1}{x}\right)^2+2< 0\)
Đặt \(x+\frac{1}{x}=a\)( \(a>2\) )
\(pt\Leftrightarrow3a-2a^2+2< 0\)
\(\Leftrightarrow2a^2-3a-2>0\)
\(\Leftrightarrow2\left(a^2-\frac{3}{2}a-1\right)>0\)
\(\Leftrightarrow2\left(a^2-2\cdot a\cdot\frac{3}{4}+\frac{9}{16}-\frac{25}{16}\right)>0\)
\(\Leftrightarrow2\left[\left(a-\frac{3}{4}\right)^2-\frac{25}{16}\right]\)
\(\Leftrightarrow2\left(a-\frac{3}{4}\right)^2-\frac{25}{8}>0\)
\(\Leftrightarrow2\left(a-\frac{3}{4}\right)^2>\frac{25}{8}\)
Ta có \(a>2\Leftrightarrow2\left(a-\frac{3}{4}\right)^2>2\left(2-\frac{3}{4}\right)^2=\frac{25}{8}\)( luôn đúng )
Vậy ta có đpcm.
