Đặt vế trái của BĐT là P:

\(P=\sqrt{\left(a+2\right)\left(b+2\right)}+\sqrt{2b.\left(a+1\right)}\)

\(P\le\dfrac{1}{2}\left(a+2+b+2\right)+\dfrac{1}{2}\left(2b+a+1\right)\)

\(P\le\dfrac{1}{2}\left(2a+3b+5\right)=\dfrac{1}{2}.2024=1012\)

Dấu "=" không xảy ra

\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)

\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)

\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)

\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)

\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)

\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)

\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)

\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)

\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)

Từ hệ phương trình \(\Rightarrow\left(\sqrt{x-2018}-\sqrt{x-2019}\right)+\left(\sqrt{y-2018}-\sqrt{y-2019}\right)=2\)

Ta có: \(\sqrt{x-2018}-\sqrt{x-2019}\le\sqrt{\left(x-2018\right)-\left(x-2019\right)}=1\) Dấu = xảy ra khi và chỉ khi x = 2019

Tương tự: \(\sqrt{y-2018}-\sqrt{y-2019}\le1\)

Dấu = xảy ra khi và chỉ khi y = 2019

Nên: \(\left(\sqrt{x-2018}-\sqrt{x-2019}\right)+\left(\sqrt{y-2018}-\sqrt{y-2019}\right)\le2\)

Dấu = xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x=2019\\y=2019\end{matrix}\right.\)

Kết luận nghiệm pt: \(\left\{{}\begin{matrix}x=2019\\y=2019\end{matrix}\right.\)

\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)

\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)

\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)