\(\Leftrightarrow\frac{x}{5}+\frac{y}{6}+\frac{z}{4}\le1\)

Đặt \(\left(\frac{x}{5};\frac{y}{6};\frac{z}{4}\right)=\left(a;b;c\right)\Rightarrow0\le a;b;c\le1\) và \(a+b+c\le1\)

\(T=25a^2+36b^2+16c^2-20a-24b-4c\)

\(25a\left(a-\frac{32}{25}\right)\le0\Rightarrow25a^2\le32a\)

\(36b\left(b-1\right)\le0\Rightarrow36b^2\le36b\)

\(16c\left(c-1\right)\le0\Rightarrow16c^2\le16c\)

\(\Rightarrow T\le32a+36b+16c-20a-24b-4c=12\left(a+b+c\right)\le12\)

\(T_{max}=12\) khi \(\left\{{}\begin{matrix}a=0\\b=0\\c=1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=0\\b=1\\c=0\end{matrix}\right.\)

\(4x^2+4y^2\ge8xy\)

\(16x^2+z^2\ge8zx\)

\(16y^2+z^2\ge8yz\)

Cộng vế với vế:

\(20x^2+20y^2+2z^2\ge8\left(xy+yz+zx\right)\)

\(\Leftrightarrow10x^2+10y^2+z^2\ge4\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)

\(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\)

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

\(\Rightarrow P\le\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}=x+y+z+3=4\)

\(P_{max}=4\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị

Chắc dùng Mincowski

\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)

\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)

\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)

\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)

\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)

\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)

\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)

\(=\sqrt{189}\)

Dấu "=" xảy ra <=> x = y = z = 4

\(P=\sqrt{y}\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}=\left(6-\sqrt{x}-\sqrt{z}\right)\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}\)

\(P=-x+6\sqrt{x}-2z+12z=-\left(\sqrt{x}-3\right)^2-2\left(\sqrt{z}-3\right)^2+27\le27\)

\(P_{max}=27\) khi \(\left(x;y;z\right)=\left(9;0;9\right)\)