Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)

Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì  \(\left(a-b=b-c\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

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

\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)

Ta có : \(xyz=1\rightarrow\left\{{}\begin{matrix}xy=\dfrac{1}{z}\\xz=\dfrac{1}{y}\\yz=\dfrac{1}{x}\end{matrix}\right.\)

Do đó : \(A=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)

\(A=1+x+y+z+xy+yz+xz+xyz\)

\(A=1+x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+1\)

\(A=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)+2\)

Áp dụng BĐT \(a+b\ge2\sqrt{ab}\left(a,b>0\right)\) 

Dấu \(=\) xảy ra \(\Leftrightarrow a=b\)

với \(x,y,z>0\) Ta được :

\(A\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}+2=2+2+2+2=8\)

Dấu \(=\) xảy ra \(\Leftrightarrow\)

\(\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\\z^2=1\end{matrix}\right.\Rightarrow x=y=z=1\) ( vì \(x,y,z>0\) )