a: \(ax+by+cz\)

\(=x^3-xyz+y^3-xyz+z^3-xyz\)

\(=x^3+y^3+z^3-3xyz\)

b: \(ax+by+cz\)

\(=x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3yxz\)

\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Ta có: \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

=>\(a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2\cdot ax\cdot by+2\cdot by\cdot cz+2\cdot ax\cdot cz\)

=>\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2\cdot ax\cdot by-2\cdot by\cdot cz-2\cdot ax\cdot cz=0\)

=>\(\left(a^2y^2-2\cdot ay\cdot bx+b^2x^2\right)+\left(b^2z^2-2\cdot bz\cdot cy+c^2y^2\right)+\left(az^2-2\cdot az\cdot cx+c^2x^2\right)=0\)

=>\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

=>ay-bx=0 và bz-cy=0 và az-cx=0

=>ay=bx và bz=cy và az=cx

=>a/x=b/y; b/y=c/z; a/x=c/z

=>a/x=b/y=c/z

\(ax+by+cz\\ =x\left(x^2-yz\right)+y\left(y^2-xz\right)+z\left(z^2-xy\right)\\ =x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Lại có \(a+b+c=x^2+y^2+z^2-xy-yz-xz\)

Vậy ta được đpcm

\(2x-2y=by+cz-cz-ax=by-ax\)

\(\Rightarrow2x-2y=by-ax\)

\(\Rightarrow2x+ax=2y+by\)

\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)

\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)

\(2z-2y=ax+by-cz-ax=by-cz\)

\(\Rightarrow2z+cz=2y+by\)

\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)

\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)

\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)

thanks bạn nha

\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)

\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)

\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)

\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)

\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)

\(\Rightarrow A\ge2\)

\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị