Ta có \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)
Xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)
<=> \(x^3\ge\left(2018^2-2.2018.x+x^2\right)\left(x-\frac{1009}{2}\right)\)
<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2018.2\right)+x\left(2018.1009+2018^2\right)-\frac{2018^2.1009}{2}\)
<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)
<=> \(\frac{9081}{2}\left(x^2-\frac{2.2018}{3}.x+\left(\frac{2018}{3}\right)^2\right)\ge0\)
<=> \(\frac{9081}{2}\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)
=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)
Khi đó \(VT\ge x-\frac{1009}{2}+y-\frac{1009}{2}+z-\frac{1009}{2}=2018-\frac{3}{2}.1009=\frac{1009}{2}\)(ĐPCM)
Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)
Ta có : \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)
xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)
<=> \(x^3\ge\left(x^2-2.2018.x+2018^2\right)\left(x-\frac{1009}{2}\right)\)
<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2.2018\right)+x\left(2018^2+1009.2018\right)-\frac{2018^2.1009}{2}\ge0\)
<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)
<=> \(\frac{9081}{2}.\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)
=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)
Khi đó \(P\ge x+y+z-\frac{3.1009}{2}=\frac{1009}{2}\)(ĐPCM)
Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)