Áp dụng BĐT Cô-si cho 3 số dương \(x^2,y^2,z^2\) , ta có:\(x^2+y^2+z^2\ge3\sqrt[3]{\left(xyz\right)^2}\)

\(\Leftrightarrow\left(xyz\right)^2\le\dfrac{\left(x^2+y^2+z^2\right)^3}{27}\) \(=\dfrac{1}{27}\)

\(\Leftrightarrow-\dfrac{1}{3\sqrt{3}}\le xyz\le\dfrac{1}{3\sqrt{3}}\)

 Vậy \(max_{xyz}=\dfrac{1}{3\sqrt{3}}\). Dấu "=" xảy ra khi \(x^2=y^2=z^2\) 

\(\Rightarrow\left(x,y,z\right)=\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)\) hoặc \(\left(\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}\right)\) và các hoán vị.

Ta sẽ chứng minh \(P_{min}=1\)

TH1: \(xyz=0\)

\(\Rightarrow x^2y^2z^2=0\Rightarrow x^4+y^4+z^4=1\)

\(P=x^2+y^2+z^2\ge\sqrt{x^4+y^4+z^4}=1\)

TH2: \(xyz\ne0\) , từ điều kiện, tồn tại 1 tam giác nhọn ABC sao cho \(\left\{{}\begin{matrix}x^2=cosA\\y^2=cosB\\z^2=cosC\end{matrix}\right.\)

\(P=cosA+cosB+cosC-\sqrt{2cosA.cosB.cosC}\)

Ta sẽ chứng minh \(cosA+cosB+cosC-\sqrt{2cosA.cosB.cosC}\ge1\)

\(\Leftrightarrow4sin\dfrac{A}{2}sin\dfrac{B}{2}sin\dfrac{C}{2}\ge\sqrt{2cosA.cosB.cosC}\)

\(\Leftrightarrow8sin^2\dfrac{A}{2}sin^2\dfrac{B}{2}sin^2\dfrac{C}{2}\ge cosA.cosB.cosC\)

\(\Leftrightarrow\dfrac{8sin^2\dfrac{A}{2}sin^2\dfrac{B}{2}sin^2\dfrac{C}{2}}{8sin\dfrac{A}{2}sin\dfrac{B}{2}sin\dfrac{C}{2}cos\dfrac{A}{2}cos\dfrac{B}{2}cos\dfrac{C}{2}}\ge cotA.cotB.cotC\)

\(\Leftrightarrow tan\dfrac{A}{2}tan\dfrac{B}{2}tan\dfrac{C}{2}\ge cotA.cotB.cotC\)

\(\Leftrightarrow tanA.tanB.tanC\ge cot\dfrac{A}{2}cot\dfrac{B}{2}cot\dfrac{C}{2}\)

\(\Leftrightarrow tanA+tanB+tanC\ge cot\dfrac{A}{2}+cot\dfrac{B}{2}+cot\dfrac{C}{2}\)

Ta có:

\(tanA+tanB=\dfrac{sin\left(A+B\right)}{cosA.cosB}=\dfrac{2sinC}{cos\left(A-B\right)-cosC}\ge\dfrac{2sinC}{1-cosC}=\dfrac{2sin\dfrac{C}{2}cos\dfrac{C}{2}}{2sin^2\dfrac{C}{2}}=cot\dfrac{C}{2}\)

Tương tự: \(tanA+tanC\ge cot\dfrac{B}{2}\) ; \(tanB+tanC\ge cot\dfrac{A}{2}\)

Cộng vế với vế ta có đpcm

Vậy \(P_{min}=1\) khi \(\left(x^2;y^2;z^2\right)=\left(1;0;0\right)\) và các hoán vị hoặc \(\left(x^2;y^2;z^2\right)=\left(\dfrac{1}{2};\dfrac{1}{2};\dfrac{1}{2}\right)\)

\(P=x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^3=\dfrac{64}{3}\)

\(P_{min}=\dfrac{64}{3}\) khi \(x=y=z=\dfrac{4}{3}\)

Đặt \(\left(x;y;z\right)=\left(a+1;b+1;c+1\right)\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c\ge0\end{matrix}\right.\)

\(\Rightarrow0\le a;b;c\le1\) \(\Rightarrow\left\{{}\begin{matrix}a^2\le a\\b^2\le b\\c^2\le c\end{matrix}\right.\) \(\Rightarrow a^2+b^2+c^2\le a+b+c=1\)

\(P=\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2\)

\(P=a^2+b^2+c^2+2\left(a+b+c\right)+3=a^2+b^2+c^2+5\le1+5=6\)

\(P_{max}=6\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị hay \(\left(x;y;z\right)=\left(1;1;2\right)\) và hoán vị