Áp dụng bđt Schwarz ta có:

\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).

\(2ab+a+b=2a^2+2b^2\ge2ab+\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\)

\(F=\dfrac{a^4}{ab}+\dfrac{b^4}{ab}+2020\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{\left(a^2+b^2\right)^2}{2ab}+\dfrac{8080}{a+b}\ge a^2+b^2+\dfrac{8080}{a+b}\)

\(F\ge\dfrac{\left(a+b\right)^2}{2}+\dfrac{8080}{a+b}=\dfrac{\left(a+b\right)^2}{2}+\dfrac{4}{a+b}+\dfrac{4}{a+b}+\dfrac{8072}{a+b}\)

\(F\ge3\sqrt[3]{\dfrac{16\left(a+b\right)^2}{\left(a+b\right)^2}}+\dfrac{8072}{2}=...\)

\(Q=\dfrac{2002}{a}+\dfrac{2017}{b}+2996a-5501b=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-\left(5012a+7518b\right)\)

\(=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-2506\left(2a+3b\right)\)

Áp dụng bất đẳng thức Cosi cho 2 số dương:

\(\left\{{}\begin{matrix}\dfrac{2002}{a}+8008\ge2\sqrt{\dfrac{2002}{a}.8008}=8008\\\dfrac{2017}{b}+2017b\ge2\sqrt{\dfrac{2017}{b}.2017b}=4034\end{matrix}\right.\)

Ta có: \(2a+3b=4\Rightarrow-\left(2a+3b\right)=-4\Leftrightarrow-2506\left(2a+3b\right)=-10024\)

\(\Rightarrow Q\ge8008+4034-10024=2018\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)

Ta thấy \(ab\le\dfrac{a^2+b^2}{2}=1\) và \(a+b\le\sqrt{2\left(a^2+b^2\right)}=2\). Áp dụng BĐT B.C.S, ta được \(P=\dfrac{a^4}{ba^2+a^2}+\dfrac{b^4}{ab^2+b^2}\) \(\ge\dfrac{\left(a^2+b^2\right)^2}{ba^2+ab^2+a^2+b^2}=\dfrac{2^2}{ab\left(a+b\right)+2}\ge\dfrac{4}{1.2+2}=1\)

ĐTXR \(\Leftrightarrow a=b=1\)

Vậy GTNN của P là 1 khi \(a=b=1\)

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

\(P=\dfrac{4ab}{a+2b}+\dfrac{9ca}{a+4c}+\dfrac{4bc}{b+c}\)

\(P=\dfrac{4abc}{ac+2bc}+\dfrac{9abc}{ab+4bc}+\dfrac{4abc}{ab+ac}\)

\(P=abc\left(\dfrac{4}{ac+2bc}+\dfrac{9}{ab+4bc}+\dfrac{4}{ab+ac}\right)\)

\(P\ge abc.\dfrac{\left(2+3+2\right)^2}{ac+2bc+ab+4bc+ab+ac}\)

\(P\ge abc.\dfrac{49}{2ab+6bc+2ca}\)

\(P\ge abc.\dfrac{49}{7abc}\) (vì \(2ab+6bc+2ca=7abc\))

\(P\ge7\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{ac+2bc}=\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\\2ab+6bc+2ca=7abc\end{matrix}\right.\)

\(\dfrac{2}{ac+2bc}=\dfrac{2}{ab+ac}\) \(\Leftrightarrow2b=a\)

Có \(\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\) 

\(\Leftrightarrow\dfrac{3}{2b^2+4bc}=\dfrac{2}{2b^2+2bc}\) 

\(\Leftrightarrow3b^2+3bc=2b^2+4bc\)

\(\Leftrightarrow b^2=bc\Leftrightarrow b=c\)

\(\Rightarrow a=2b=2c\)

Lại có \(2ab+6bc+2ca=7abc\) \(\Rightarrow4b^2+6b^2+4b^2=14b^3\)

\(\Leftrightarrow b=1\)

\(\Leftrightarrow\left(a,b,c\right)=\left(2,1,1\right)\)

Vậy \(min_P=7\)