Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)

Đơn giản là kiên nhẫn tính toán và tách biểu thức:

\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)

Sau đó Cô-si cho từng ngoặc là được

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=2(\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2})+\frac{1}{2(ab+bc+ac)}\\ \geq 2.\frac{9}{2(ab+bc+ac)+a^2+b^2+c^2}+\frac{1}{2(ab+bc+ac)}\\ =\frac{18}{(a+b+c)^2}+\frac{1}{2(ab+bc+ac)}\\ =18+\frac{1}{2(ab+bc+ac)}\)

Áp dụng BĐT AM-GM:

$2(ab+bc+ac)\leq 2.\frac{(a+b+c)^2}{3}=\frac{2}{3}$

$\Rightarrow \frac{1}{2(ab+bc+ac)}\geq \frac{3}{2}$

$\Rightarrow P\geq 18+\frac{3}{2}=\frac{39}{2}$
Vậậy $P_{\min}=\frac{39}{2}$ khi $a=b=c=\frac{1}{3}$

\(Q=ac+bc-2022ab\le ac+bc=c\left(a+b\right)\le\dfrac{1}{4}\left(c+a+b\right)^2=\dfrac{1}{4}\)

\(Q_{max}=\dfrac{1}{4}\) khi \(\left\{{}\begin{matrix}a+b+c=1\\ab=0\\c=a+b\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;\dfrac{1}{2};\dfrac{1}{2}\right);\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)

\(Q=c\left(a+b\right)-2022ab\ge c\left(a+b\right)-\dfrac{1011}{2}\left(a+b\right)^2\)

\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}\left(1-c\right)^2\)

\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}c\left(c-2\right)-\dfrac{1011}{2}\)

\(Q\ge\dfrac{c\left(1011+1013\left(1-c\right)\right)}{2}-\dfrac{1011}{2}\ge-\dfrac{1011}{2}\)

\(Q_{min}=-\dfrac{1011}{2}\) khi \(\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)