1999 is the last year of the 20th century, the last year of the 21st century is the year

We have : \(2018.4=\left(1^2+1^2+1^2+1^2\right)\left(a^4+b^4+c^4+d^4\right)\ge\left(a^2+b^2+c^2+d^2\right)^2\) 

(Inquality Bunyakovsky)

\(\Leftrightarrow-\sqrt{2018.4}\le a^2+b^2+c^2+d^2\le\sqrt{2018.4}\)

\(\Leftrightarrow0\le a^2+b^2+c^2+d^2< 90\)

Edit post \(\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)

Have : 

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\Rightarrow\dfrac{1}{x}\ge\left(1-\dfrac{1}{y}\right)+\left(1-\dfrac{1}{z}\right)=\dfrac{y-1}{y}\)

\(\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\) (Cauchy)

Similar \(\left\{{}\begin{matrix}\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{xz}}\\\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(x-1\right)}{xy}}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{xyz}\ge8\sqrt{\dfrac{\left(x-1\right)^2\left(y-1\right)^2\left(z-1\right)^2}{\left(xyz\right)^2}}=\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)

\(\Rightarrow1\ge8\left(x-1\right)\left(y-1\right)\left(z-1\right)\Rightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)

We have : \(A=x^2-xy+y^2=\left(x^2-xy+\dfrac{1}{4}y^2\right)+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x;y;A\ne0\)

 \(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\left(\dfrac{4}{5}x+\dfrac{6}{5}x\right)+\left(\dfrac{4}{5}y+\dfrac{1}{5}y\right)+\dfrac{30}{x}+\dfrac{5}{y}\)

\(=\left(\dfrac{4}{5}x+\dfrac{4}{5}y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\)

\(\dfrac{4}{5}\left(x+y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\)

Because x; y is positive we have :

\(\dfrac{6x}{5}+\dfrac{30}{x}\ge2\sqrt{\dfrac{6x}{5}.\dfrac{30}{x}}=2\sqrt{36}=12\) (Cauchy)

\(\dfrac{y}{5}+\dfrac{5}{y}\ge2\sqrt{\dfrac{y}{5}.\dfrac{5}{y}}=2\)(Caychy)

\(\Rightarrow\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\ge14\)

We have \(x+y\ge10\Rightarrow\dfrac{4}{5}\left(x+y\right)\ge\dfrac{4}{5}.10=8\)

\(\Rightarrow\dfrac{4}{5}\left(x+y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\ge14+8=22\)

The equation occurs when \(x=y=5\)

\(P_{min}=22\Leftrightarrow x=y=5\)