Dự đoán dấu = xảy ra khi x=y=\(\dfrac{z}{2}\)
ta có: \(VT=3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{y^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{x^2}\)
\(=3+\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\right)+\left(\dfrac{z^2}{y^2}+\dfrac{z^2}{x^2}\right)\)
Áp dụng BĐT AM-GM: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\ge2\)
Áp dụng BĐT bunyakovsky:\(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\ge\dfrac{1}{2}\left(\dfrac{y}{z}+\dfrac{x}{z}\right)^2=\dfrac{1}{2}.\dfrac{\left(x+y\right)^2}{z^2}\)
\(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\ge\dfrac{1}{2}\left(\dfrac{z}{x}+\dfrac{z}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4z}{x+y}\right)^2=\dfrac{8z^2}{\left(x+y\right)^2}\)(AM-GM)
do đó \(VT\ge5+\dfrac{1}{2}\dfrac{\left(x+y\right)^2}{z^2}+\dfrac{8z^2}{\left(x+y\right)^2}\)
Đặt \(\dfrac{z}{x+y}=a\)(a>0)thì \(a\ge1\)do \(z\ge x+y\)
\(VT\ge8a^2+\dfrac{1}{2a^2}+5=\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{15}{2}a^2+5\ge\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{25}{2}\)
Áp dụng BĐT AM-GM: \(\dfrac{a^2}{2}+\dfrac{1}{2a^2}\ge2\sqrt{\dfrac{a^2}{4a^2}}=1\)
do đó \(VT\ge1+\dfrac{25}{2}=\dfrac{27}{2}\)(đpcm)
Dấu = xảy ra khi a=1 hay \(x=y=\dfrac{z}{2}\)
