\(A=\frac{m^2+7m+14}{\left(m+2\right)^2}\Rightarrow A\left(m+2\right)^2=m^2+7m+14\)

\(\Leftrightarrow\left(A-1\right)m^2+\left(4A-7\right)m+4A-14=0\)

- \(A-1=0\Leftrightarrow A=1\): \(m=\frac{-10}{3}\).

- \(A-1\ne0\): \(\Delta=\left(4A-7\right)^2-4\left(4A-14\right)\left(A-1\right)=16A-7\)

để phương trình có nghiệm thì \(\Delta\ge0\Leftrightarrow A\ge\frac{7}{16}\).

Vậy \(minA=\frac{7}{16}\).

Theo giả thiết, ta có: \(\frac{x}{1+x}+\frac{2y}{1+y}=1\Leftrightarrow\frac{2y}{1+y}=1-\frac{x}{1+x}=\frac{1}{x+1}\)\(\Leftrightarrow2y\left(x+1\right)=y+1\Leftrightarrow2xy^2=-y^2+y=-\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow xy^2\le\frac{1}{8}\)

Đẳng thức xảy ra khi \(x=y=\frac{1}{2}\)

\(\text{⋄}\)Xét xyz = 0 thì dễ có x = y = z = 0 (Nếu giả sử x = 0 thì 4y²(1 - x) = 0 hay y = 0 do đó 4z²(1 -  y) = 0 suy ra z = 0, tương tự đối với y, z = 0)

\(\text{⋄}\)Xét \(xyz\ne0\)thì từ hệ suy ra \(xyz=64x^2y^2z^2\left(1-x\right)\left(1-y\right)\left(1-z\right)\Leftrightarrow64xyz\left(1-x\right)\left(1-y\right)\left(1-z\right)=1\)(*)

Dễ có: \(\left(2x-1\right)^2\ge0\Leftrightarrow4x\left(1-x\right)\le1\), tương tự: \(4y\left(1-y\right)\le1;4z\left(1-z\right)\le1\)suy ra \(64xyz\left(1-x\right)\left(1-y\right)\left(1-z\right)\le1\)

Như vậy điều kiện để (*) xảy ra là \(x=y=z=\frac{1}{2}\)

Vậy hệ có 2 nghiệm \(\left(x,y,z\right)\in\left\{\left(0;0;0\right),\left(\frac{1}{2};\frac{1}{2};\frac{1}{2}\right)\right\}\)

Ta có: \(\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\)\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)

Ta luôn có: \(\left(\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right)^2\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2+2.\)\(\left(\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\right)\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2\ge2\)(*)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+1+\left(\frac{b+c}{b-c}\right)^2+1+\left(\frac{c+a}{c-a}\right)^2+1\ge5\)

\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{\left(a-b\right)^2}+\frac{2\left(b^2+c^2\right)}{\left(b-c\right)^2}+\frac{2\left(c^2+a^2\right)}{\left(c-a\right)^2}\ge5\)\(\Leftrightarrow\frac{a^2+b^2}{\left(a-b\right)^2}+\frac{b^2+c^2}{\left(b-c\right)^2}+\frac{c^2+a^2}{\left(c-a\right)^2}\ge\frac{5}{2}\)(1)

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

Lấy (1) + (2), ta được: \(\frac{a^2+ab+b^2}{\left(a-b\right)^2}+\frac{b^2+bc+c^2}{\left(b-c\right)^2}+\frac{c^2+ca+a^2}{\left(c-a\right)^2}\ge\frac{9}{4}\)

\(\Leftrightarrow\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)(đpcm)

Chú ý: Từ đây ta có thể biến thành một BĐT khác khó hơn: \(\frac{a^3+b^3}{\left(a-b\right)^3}+\frac{b^3+c^3}{\left(b-c\right)^3}+\frac{c^3+a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)

Ta có: \(\sqrt{2x^2-4x+5}=\sqrt{2x^2-4x+2+3}=\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)

Lại có: \(\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(\sqrt{2}x-\sqrt{2}\right)^2+3\ge3\)

\(\Rightarrow\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge\sqrt{3}\)

Vậy Min y là \(2+\sqrt{3}\)

\(y=2+\sqrt{2x^2-4x+5}=2+\sqrt{2x^2-4x+2+3}\)

\(=2+\sqrt{2\left(x^2-2x+1\right)+3}=2+\sqrt{2\left(x-1\right)^2+3}\)

Vì \(\left(x-1\right)^2\ge0\)\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2\ge0\)\(\forall x\)\(\Rightarrow2\left(x-1\right)^2+3\ge3\)\(\forall x\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\)\(\forall x\)

\(\Rightarrow y=2+\sqrt{2\left(x-1\right)^2+3}\ge2+\sqrt{3}\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(miny=2+\sqrt{3}\)\(\Leftrightarrow x=1\)