\(P=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)+1}{a+b+c-abc}=\dfrac{\left(a+b+c\right)^2+1}{a+b+c-abc}\ge\dfrac{\left(a+b+c\right)^2+1}{a+b+c}\)

\(\Rightarrow P\ge a+b+c+\dfrac{1}{a+b+c}\) (1)

\(P=\dfrac{a^2+b^2+c^2+3\left(ab+bc+ca\right)}{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}=\dfrac{\left(a+b\right)\left(b+c\right)+\left(b+c\right)\left(c+a\right)+\left(a+b\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(P=\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{a+b+c}\left(\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+c+b}{a+c}\right)\)

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

\(P\ge\dfrac{1}{a+b+c}\left(3+\dfrac{\left(a+b+c\right)^2}{2}\right)=\dfrac{3}{a+b+c}+\dfrac{a+b+c}{2}\)

\(\Rightarrow3P\ge\dfrac{3}{2}\left(a+b+c\right)+\dfrac{9}{a+b+c}\) (2)

Cộng vế (1) và (2):

\(\Rightarrow4P\ge\dfrac{5}{2}\left(a+b+c\right)+\dfrac{10}{a+b+c}\ge2\sqrt{\dfrac{50\left(a+b+c\right)}{2\left(a+b+c\right)}}=10\)

\(\Rightarrow P\ge\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;1;0\right)\) và các hoán vị

Ta có: \(\sqrt{a^2+b^2+c^2}\ge\sqrt{\dfrac{\left(a+b+c\right)^2}{3}}=\sqrt{3};\sqrt{a^2+b^2+c^2}\le\sqrt{\left(a+b+c\right)^2}=3\).

Đặt \(\sqrt{a^2+b^2+c^2}=t\) \((\sqrt{3}\leq t\leq 3)\).

Ta có: \(P=t+\dfrac{9-t^2}{4}+\dfrac{1}{t^2}=\dfrac{4t^3+9t^2-t^4+4}{4t^2}\).

\(\Rightarrow P-\dfrac{28}{9}=\dfrac{\left(3-t\right)\left(9t^3-9t^2+4t+12\right)}{36}\).

Do \(\sqrt{3}\le t\le3\) nên \(3-t\geq 0\); \(9t^3-9t^2+4t+12>4t+12>0\).

Nên \(P\ge\dfrac{28}{9}\).

Đẳng thức xảy ra khi t = 3, tức (a, b, c) = (0; 0; 3) và các hoán vị.

Vậy...

\(P=\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{\left(a+b+c\right)^3}{abc}\)

\(\ge\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{9\left(a+b+c\right)^2}{ab+bc+ca}\)

\(=\left[\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\right]+\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+18\)

\(\ge2+8+18=28\)

Đẳng thức xảy ra khi \(a=b=c\)

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

\(\Rightarrow2P+1=\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{2\left(a+b+c\right)}{abc}\right)=\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\right)\)

\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{18}{ab+bc+ca}\right)\)

\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{16}{ab+bc+ca}\right)\)

\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{16}{ab+bc+ca}\right)\)

\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{9}{\left(a+b+c\right)^2}+\frac{48}{\left(a+b+c\right)^2}\right)=57\)

\(\Rightarrow P\ge28\)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng bất đẳng thức AM - GM ta có:

\(\dfrac{\sqrt{a^2+b^2+c^2}}{8}+\dfrac{\sqrt{a^2+b^2+c^2}}{8}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{3}{4}\). (1)

Đặt \(\sqrt{a^2+b^2+c^2}=t\Rightarrow\sqrt{\dfrac{4}{3}}\le t\le2\).

\(\dfrac{3\sqrt{a^2+b^2+c^2}}{4}+\dfrac{ab+bc+ca}{2}=\dfrac{3t}{4}+\dfrac{4-2t^2}{4}=\dfrac{\left(2-t\right)\left(2t+1\right)}{4}+\dfrac{3}{2}\ge\dfrac{3}{2}\). (2)

Cộng vế với vế của (1), (2) ta được \(P\ge\dfrac{9}{4}\).

...

\(M\ge3\left(ab+bc+ca\right)+2\sqrt{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}=3\left(ab+bc+ca\right)+2\sqrt{1-2\left(ab+bc+ca\right)}\)

\(\text{Đặt }t=\sqrt{1-2\left(ab+bc+ca\right)}\Rightarrow ab+bc+ca=\frac{1-t^2}{2}\)

\(\text{Ta có: }0\le ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)

\(\Rightarrow ab+bc+ca\in\left[0;\frac{1}{3}\right]\)

\(\Rightarrow-2\left(ab+bc+ca\right)\in\left[-\frac{2}{3};0\right]\)

\(\Rightarrow1-2\left(ab+bc+ca\right)\in\left[\frac{1}{3};1\right]\)

\(\Rightarrow t\in\left[\frac{1}{\sqrt{3}};1\right]\)

\(M=3.\frac{1-t^2}{2}+2t=-\frac{3}{2}t^2+2t+\frac{3}{2}\)

Lập bảng biến thiên hàm bậc 2, suy ra \(\text{Min }M\text{ (}t\in\left[\frac{1}{\sqrt{3}};1\right]\text{) }=2\text{ tại }t=1\)

Vậy GTNN của M là 2 khi t = 1 hay \(ab+bc+ca=0\Leftrightarrow\left(a;b;c\right)=\left(1;0;0\right);\left(0;0;1\right);\left(0;1;0\right)\)

\(\dfrac{a^3}{a^2+bc}=a-\dfrac{abc}{a^2+bc}\ge a-\dfrac{abc}{2a\sqrt{bc}}=a-\dfrac{\sqrt{bc}}{2}\)

\(\dfrac{b^3}{b^2+ca}\ge b-\dfrac{\sqrt{ac}}{2};\dfrac{c^3}{c^2+ab}\ge c-\dfrac{\sqrt{ab}}{2}\)

\(\Rightarrow M\ge a+b+c-\left(\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{bc}}{2}+\dfrac{\sqrt{ca}}{2}\right)=2022-\left(\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\right)\)

\(do:\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)

\(\Rightarrow M\ge2022-\dfrac{a+b+c}{2}=2022-\dfrac{2022}{2}=1011\)

\(min_M=2021\Leftrightarrow a=b=c=674\)

có đoạn  bạn sửa lại tí nhé tại lúc đầu mình đọc đề thành \(a+b+c=2022\)

\(M\ge a+b+c-\left(\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\right)\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\ge\dfrac{2022}{2}=1011\)

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\ge\frac{9}{ab+ac+bc}\right)\)

\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}\ge\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}\right)-\frac{3}{2}\left(1\right)\)

Lại có: \(\frac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2+2\left(ab+bc+ac\right)}{30\left(a^2+b^2+c^2\right)}\)

\(=\frac{1}{30}+\frac{1}{15}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow P=\frac{1}{30}-\frac{3}{2}+\frac{1}{5}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)+\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)

\(=\frac{1}{15}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}+\frac{ab+bc+ca}{a^2+b^2+c^2}-22\right)\ge-\frac{4}{3}\)