dk \(x\ge-1\)

dat \(x=a\ge-1,\sqrt{x+1}=b\ge0\)

=> \(\hept{\begin{cases}a^2+2b^2=3ab.\left(1\right)\\b^2-a=1.\left(2\right)\end{cases}}\) 

xet a=0 => x=0 ko phai la nghiem pt =>

(1) <=> \(1+2\left(\frac{b}{a}\right)^2-3\frac{b}{a}=0\Leftrightarrow\orbr{\begin{cases}\frac{b}{a}=1\\\frac{b}{a}=\frac{1}{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}b=a\\a=2b\end{cases}}\)

th b=a thay vao (2) => \(b^2-b-1=0\Rightarrow b=\frac{1+\sqrt{5}}{2}\)=a=x (tmdk)

th a=2b thay vao (2) => \(b^2-2b-1=0\Rightarrow b=1+\sqrt{2}=>x=a=2+2\sqrt{2}\)(tmdk)

Vay \(S=\left\{2+2\sqrt{2};\frac{1+\sqrt{5}}{2}\right\}\)

\(P=\frac{3a+3b+2c}{\sqrt{6\left(a^2+5\right)}+\sqrt{6\left(b^2+5\right)}+\sqrt{c^2+5}}\)

\(=\frac{3a+3b+2c}{\sqrt{6\left(a^2+ab+bc+ca\right)}+\sqrt{6\left(b^2+ab+bc+ca\right)}+\sqrt{c^2+ab+bc+ca}}\)(Do ab + bc + ca = 5)

\(=\frac{3a+3b+2c}{\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}}\)

Áp dụng BĐT AM - GM, ta được:

\(\sqrt{6\left(a+b\right)\left(a+c\right)}=2\sqrt{\frac{6}{4}\left(a+b\right)\left(a+c\right)}\)\(\le\frac{6}{4}\left(a+b\right)+\left(a+c\right)=\frac{5}{2}a+\frac{6}{4}b+c\)

\(\sqrt{6\left(b+a\right)\left(b+c\right)}=2\sqrt{\frac{6}{4}\left(b+a\right)\left(b+c\right)}\)\(\le\frac{6}{4}\left(a+b\right)+\left(b+c\right)=\frac{6}{4}a+\frac{5}{2}b+c\)

\(\sqrt{\left(c+a\right)\left(c+b\right)}\le\frac{\left(c+a\right)+\left(c+b\right)}{2}=c+\frac{a}{2}+\frac{b}{2}\)

Cộng theo vế của 3 BĐT trên, ta được: \(\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\)\(\le\frac{9}{2}a+\frac{9}{2}b+3c\)

\(\Rightarrow\frac{3a+3b+2c}{\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}}\)\(\ge\frac{3a+3b+2c}{\frac{9}{2}a+\frac{9}{2}b+3c}=\frac{2}{3}\)

Đẳng thức xảy ra khi \(a=b=1;c=2\)