\(4\le\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\le\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}+2\right)^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow2\le\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\Rightarrow x+y\ge2\)

\(\Rightarrow P\ge\dfrac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Dấu "=" xảy ra khi \(x=y=1\)

Trước hết áp dụng BĐT: \(ab\le\dfrac{1}{4}\left(a+b\right)^2\)

Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\le\dfrac{1}{4}\left(\sqrt{x}+1+\sqrt{y}+1\right)^2\)

Mà \(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\ge4\Rightarrow\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}+2\right)^2\ge4\)

\(\Rightarrow\left(\sqrt{x}+\sqrt{y}+2\right)^2\ge4^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\ge2\)

Lại áp dụng tiếp: \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Ta được: \(\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\)

\(\Rightarrow\sqrt{2\left(x+y\right)}\ge\sqrt{x}+\sqrt{y}\ge2\)

Bình phương lên: \(2\left(x+y\right)\ge4\Rightarrow x+y\ge2\)

Phần cuối chắc là hoàn toàn cơ bản rồi

\(VT=\sqrt{\left(\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\right)^2}\)

\(\le\sqrt{3\left(x+y+z+3\right)}=\sqrt{\left[9-2\left(x+y+z\right)\right]+5\left(x+y+z\right)}\)

\(=\sqrt{\left[9-\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]+5\left(x+y+z\right)}\le\sqrt{5\left(x+y+z\right)}=VP\)

Đẳng thức xảy ra khi \(a=b=c=\frac{3}{2}\)

Theo giả thiết \(2=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\Rightarrow x+y+z\ge\frac{9}{2}\)

\(\Rightarrow\frac{2}{3}\left(x+y+z\right)\ge3\)

\(VT=\sqrt{\left(\Sigma_{cyc}\frac{\sqrt{x+1}}{\sqrt{5\left(x+y+z\right)}}.\sqrt{5\left(x+y+z\right)}\right)^2}\le\sqrt{15\left(x+y+z\right)\left[\Sigma_{cyc}\frac{x+1}{5\left(x+y+z\right)}\right]}\)

\(=\sqrt{3\left(x+y+z+3\right)}\le\sqrt{3\left(x+y+z+\frac{2}{3}\left(x+y+z\right)\right)}=\sqrt{5\left(x+y+z\right)}=VP\)