\(x^2+\dfrac{1}{x^2}=\dfrac{x^2}{16}+\dfrac{1}{x^2}+\dfrac{15x^2}{16}\)
áp dụng bđt cô si vs 2 số ko âm \(\dfrac{x^2}{16},\dfrac{1}{x^2}\)
ta có \(\dfrac{x^2}{16}+\dfrac{1}{x^2}\ge\dfrac{1}{2}\)
=> \(x^2+\dfrac{1}{x^2}\ge\dfrac{1}{2}+\dfrac{15x^2}{16}\) mà x ≥2
=> \(x^2+\dfrac{1}{x^2}\ge\dfrac{1}{2}+\dfrac{15.2^2}{16}\ge\dfrac{17}{4}\)
dấu "=" xảy ra khi x = 2