Tổng các số trong phương trình là 1, vì vậy ta có: 3a + 2b + c = 1.

Với số tự nhiên a, b và c, ta có thể thử các giá trị để tìm bộ ba số thỏa mãn phương trình.

Ví dụ, ta có thể thử a = 1, b = 1 và c = -4, thì 3a + 2b + c = 3 + 2 + (-4) = 1, phương trình được thỏa mãn.

Vậy, một bộ ba số tự nhiên khác 0 thỏa mãn phương trình đã cho là a = 1, b = 1 và c = -4.

Áp dụng BĐT BSC:

\(P=\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\ge\dfrac{\left(1+2+3\right)^2}{a+b+c}=\dfrac{36}{1}=36\)

\(minP=36\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{3}\\c=\dfrac{1}{2}\end{matrix}\right.\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

BĐT trở thành:

\(\dfrac{y^2}{xz}+\dfrac{z^2}{xy}+\dfrac{x^2}{yz}\ge\dfrac{3}{2}\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}-1\right)\)

\(\Leftrightarrow2\left(x^3+y^3+z^3\right)+3xyz\ge3x^2y+3y^2z+3z^2x\)

Áp dụng BĐT Schur ta có:

\(x^3+y^3+z^3+3xyz\ge x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\)

\(\Rightarrow VT\ge\left(x^3+xy^2\right)+\left(y^3+yz^2\right)+\left(z^3+zx^2\right)+x^2y+y^2z+z^2x\ge3\left(x^2y+y^2z+z^2x\right)\)

Sửa \(\le\) thành \(\ge\) nha bạn

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

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

Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)

Áp dụng BĐT cosi:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)

\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)

\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)

Cộng VTV:

\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu \("="\Leftrightarrow a=b=c=3\)

Không mất tính tổng quát, giả sử \(a\le b\le c\)

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

\(\Rightarrow2=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}=\dfrac{3}{a}\)

\(\Rightarrow a\le\dfrac{3}{2}\)

Mà a là số nguyên dương

\(\Rightarrow a=1\)

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Rightarrow\dfrac{1}{b}+\dfrac{1}{c}=1\le\dfrac{1}{b}+\dfrac{1}{b}=\dfrac{2}{b}\)

\(\Rightarrow b\le2\)

\(\Rightarrow y\in\left\{1;2\right\}\)

\(\Rightarrow z\in\left\{1;2\right\}\)

Vậy \(\left(x;y;z\right)\in\left\{\left(1;2;2\right),\left(2;2;1\right),\left(2;1;2\right),\left(2;2;1\right)\right\}\)

Nhân vế với vế của giả thiết:

\(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow\left(ab+\dfrac{1}{bc}+\dfrac{a}{c}+1\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow abc+\dfrac{1}{abc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c=\dfrac{28}{3}\) (1)

Cộng vế với vế giả thiết:

\(\Rightarrow a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4+1+\dfrac{7}{3}=\dfrac{22}{3}\) (2)

(1);(2) \(\Rightarrow abc+\dfrac{1}{abc}+\dfrac{22}{3}=\dfrac{28}{3}\)

\(\Rightarrow abc+\dfrac{1}{abc}=2\)

\(\Rightarrow\left(abc\right)^2-2\left(abc\right)+1=0\)

\(\Rightarrow\left(abc-1\right)^2=0\)

\(\Rightarrow abc=1\)

Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\Leftrightarrow a^2+2ab+b^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\right)\)

\(\dfrac{1}{2a+b+c}=\dfrac{1}{4}.\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)

CMTT \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\\\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\end{matrix}\right.\)

\(\Rightarrow M=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{2}{2a}+\dfrac{2}{2b}+\dfrac{2}{2c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}.4=1\)

\(minM=1\Leftrightarrow a=b=c=\dfrac{3}{4}\)