Đặt \(\left(x;2y;3z\right)=\left(a;b;c\right)\Rightarrow a+b+c=6\)

\(P=\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\)

Ta có đánh giá: \(\frac{5b^3-a^3}{ab+3b^2}\le2b-a\)

Thật vậy, BĐT tương đương: \(5b^3-a^3\le\left(2b-a\right)\left(ab+3b^2\right)\)

\(\Leftrightarrow a^3-a^2b-ab^2+b^3\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)

Tương tự: \(\frac{5c^3-b^3}{bc+3c^2}\le2c-b\) ; \(\frac{5a^3-c^3}{ca+3a^2}\le2a-c\)

Cộng vế với vế: \(P\le a+b+c=6\)

\(P_{max}=6\) khi \(a=b=c=2\) hay \(\left(x;y;z\right)=\left(2;1;\frac{2}{3}\right)\)

UCT mở rộng: ta sẽ đi tìm m;n sao cho: \(\frac{5b^3-a^3}{ab+3b^2}\le ma+nb\)

\(\Leftrightarrow a^3+ma^2b+\left(3m+n\right)ab^2+\left(3n-5\right)b^3\ge0\) (1)

\(\Leftrightarrow x^3+m.x^2+\left(3m+n\right)x+\left(3n-5\right)\ge0\) với \(x=\frac{a}{b}\)

Dự đoán rằng sẽ phân tích về dạng \(\left(a-b\right)^2.P\left(a;b\right)\) hay \(\left(x-1\right)^2P\left(x\right)\)

Do đó (1) phải có nghiệm \(x=1\)

\(\Rightarrow4m+4n-4=0\Rightarrow n=1-m\)

Thay vào: \(x^3+mx^2+\left(2m+1\right)x-3m-2\ge0\)

Hoocne hạ bậc: \(\left(x-1\right)\left(x^2+\left(m+1\right)x+3m+2\right)\ge0\)

\(\Rightarrow x^2+\left(m+1\right)x+3m+2\) cũng có 1 nghiệm \(x=1\)

\(\Rightarrow4m+4=0\Rightarrow m=-1\Rightarrow n=2\)

Ta có: \(x+2y+3x=0\Leftrightarrow x=-\left(2y+3z\right)\)

Lại có: \(2xy+6yz+3xz=0\Leftrightarrow x\left(2y+3z\right)+6yz=0\)

\(\Leftrightarrow-\left(2y+3z\right)\left(2y+3z\right)+6yz=0\Leftrightarrow-\left(2y+3z\right)^2+6yz=0\)

\(\Leftrightarrow\left(2y+3z\right)^2-6yz=0\Leftrightarrow4y^2+12yz+9z^2-6yz=0\)

\(\Leftrightarrow4y^2+6yz+9z^2=0\Leftrightarrow\left(2y+\dfrac{3z}{2}\right)^2+\dfrac{27z^2}{4}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+\dfrac{3z}{2}\right)^2=0\\\dfrac{27z^2}{4}=0\end{matrix}\right.\) \(\Rightarrow y=z=0\Rightarrow x=0\)

\(\Rightarrow S=\dfrac{\left(-1\right)^{2019}-1^{2017}+\left(-1\right)^{2015}}{1^{2018}+2.0^{2016}+0^{2014}+2}=\dfrac{-1-1+-1}{1+0+0+2}=\dfrac{-3}{3}=-1\)

\(\hept{\begin{cases}a=x\\b=2y\\c=3z\end{cases}}\Rightarrow a+b+c=3\)

\(Q=\frac{11b^3-a^3}{ab+4b^2}+\frac{11c^3-b^3}{bc+4c^2}+\frac{11a^3-c^3}{ca+4a^2}\)

Cần tìm \(\beta;\gamma\) sau cho \(\frac{11b^3-a^3}{ab+4b^2}\le\gamma b+\beta a\)

\(\Leftrightarrow\frac{11.\left(\frac{b}{a}\right)^3-1}{\frac{b}{a}+4\left(\frac{b}{a}\right)^2}\le\gamma\frac{b}{a}+\beta\)

\(\Leftrightarrow\frac{11t^3-1}{t+4t^2}\le\gamma t+\beta\text{ }\left(t=\frac{b}{a}\right)\)

Dự đoán Q max khi a = b = c nên t = 1;

Tới đây dùng pp hệ số bất định để tìm ra \(\gamma=3;\text{ }\beta=-1\)

Vậy ta cần chứng minh \(\frac{11b^3-a^3}{ab+4b^2}\le3b-a\Leftrightarrow-\frac{\left(a+b\right)\left(a-b\right)^2}{ab+4b^2}\le0\)

Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:

\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Chứng minh tương tự tạ có:

\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)