Đặt \(\dfrac{x}{y}=a\Rightarrow\dfrac{y}{x}=\dfrac{1}{a}\)
Viết lại BĐT, ta được:
\(3\left(a+\dfrac{1}{a}\right)-\left(a^2+\dfrac{1}{a^2}\right)\le4\)
\(\Leftrightarrow4-3\left(a+\dfrac{1}{a}\right)+\left(a^2+\dfrac{1}{a^2}\right)\ge0\)
\(\Leftrightarrow4-3a-\dfrac{3}{a}+a^2+\dfrac{1}{a^2}\ge0\)
\(\Leftrightarrow a^2-3a+2+\dfrac{1}{a^2}-\dfrac{3}{a}+2\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{a}-2\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+\dfrac{1-a}{a}.\dfrac{1-2a}{a}\ge0\)
\(\Leftrightarrow\left(a-1\right)\left[a-2+\dfrac{2a-1}{a^2}\right]\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a^3-2a^2+2a-1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a^3-a^2+a-a^2+a-1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)\left[a^2\left(a-1\right)-a\left(a-1\right)+a-1\right]\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a-1\right)\left(a^2-a+1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left[\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\ge0\) ( luôn đúng)
Dấu " = " xảy ra khi: \(a=1\Leftrightarrow x=y\)
