\(z\ge x+y\Rightarrow\dfrac{z}{x+y}\ge1\)
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P\ge\left[\dfrac{1}{2}\left(x+y\right)^2+z^2\right]\left[\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2+\dfrac{1}{z^2}\right]\)
\(P\ge\left[\dfrac{1}{2}\left(x+y\right)^2+z^2\right]\left[\dfrac{1}{2}\left(\dfrac{4}{x+y}\right)^2+\dfrac{1}{z^2}\right]=\left[\dfrac{1}{2}\left(x+y\right)^2+z^2\right]\left[\dfrac{8}{\left(x+y\right)^2}+\dfrac{1}{z^2}\right]\)
\(P\ge5+\dfrac{1}{2}.\left(\dfrac{x+y}{z}\right)^2+8.\left(\dfrac{z}{x+y}\right)^2\)
\(P\ge5+\dfrac{1}{2}\left(\dfrac{x+y}{z}\right)^2+\dfrac{1}{2}\left(\dfrac{z}{x+y}\right)^2+\dfrac{15}{2}.\left(\dfrac{z}{x+y}\right)^2\)
\(P\ge5+\dfrac{1}{2}.2\sqrt{\left(\dfrac{x+y}{z}\right)^2\left(\dfrac{z}{x+y}\right)^2}+\dfrac{15}{2}.1^2=\dfrac{27}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{z}{2}\)
