\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)

\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)

Cộng vế:

\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)

\(xy+yz+zx-xyz=1-x-y-z+xy+yz+zx-xyz\)

\(=\left(1-x\right)-y\left(1-x\right)-z\left(1-x\right)+yz\left(1-x\right)\)

\(=\left(1-x\right)\left(1-y-z+yz\right)=\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

\(xy+yz+zx+xyz+2=1+x+y+z+xy+yz+zx+xyz\)

\(=\left(1+x\right)+y\left(1+x\right)+z\left(1+x\right)+yz\left(1+x\right)\)

\(=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)

\(1+x+y+z=1+1\Rightarrow1+x=\left(1-y\right)+\left(1-z\right)\ge2\sqrt{\left(1-y\right)\left(1-z\right)}\)

Tương tự ta cũng có: \(1+y\ge2\sqrt{\left(1-z\right)\left(1-x\right)}\)

\(1+z\ge2\sqrt{\left(1-x\right)\left(1-y\right)}\)

Vậy \(S\le\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{8\left(1-x\right)\left(1-y\right)\left(1-z\right)}=\frac{1}{8}\)

Áp dụng AM - GM:

\(2x^2+\frac{1}{2}z^2\ge2\sqrt{2x^2.\frac{1}{2}z^2}=2xz\)

\(2y^2+\frac{1}{2}z^2\ge2\sqrt{2y^2.\frac{1}{2}z^2}=2yz\)(x,y,z dương)

\(x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)

Cộng từng vế của các BĐT trên:

\(T\ge2\left(xy+yz+xz\right)=10\)

(Dấu "="\(\Leftrightarrow x=1;y=1;z=2\))

Có \(3z^2\)ko ạ ?

Áp dụng bđt AM-GM ta có:

\(\hept{\begin{cases}x^2+y^2\ge2\sqrt{x^2y^2}=2xy\\y^2+z^2\ge2\sqrt{y^2z^2}=2yz\\z^2+x^2\ge2\sqrt{z^2x^2}=2zx\end{cases}}\)

Cộng theo vế các bđt trên ta được:

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)

\(\Rightarrow x^2+y^2+z^2\ge5\)

\(\Rightarrow T\ge15\)

Dấu"="xảy ra \(\Leftrightarrow x^2=y^2;y^2=z^2;z^2=x^2;xy+yz+zx=5\)

                       \(\Leftrightarrow x=y=z=\sqrt{\frac{5}{3}}\)

Vậy \(T_{min}=15\)\(\Leftrightarrow x=y=z=\sqrt{\frac{5}{3}}\)

\(A=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\)

\(\ge\frac{9}{3+x^2+y^2+z^2}\ge\frac{9}{3+3}=\frac{3}{2}\)

đoạn lớn hơn hoặc bằng cụm 9/ (3+xy+yz+zx) ấy, làm sao để có, mình ko hiểu lắm

Cái đó là bất đẳng thức cosi swat đó

Áp dụng bđt Svacsơ ta có :

\(P=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{x^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)

ta lại có : \(\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)\ge\left(xy+yz+zx\right)^2\)( bunhiacopxki )

\(\Rightarrow x^2+y^2+z^2\ge\left|xy+yz+xz\right|\ge xy+yz+xz\)

\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3zx\)

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)=3\)

\(\Rightarrow x+y+z\ge\sqrt{3}\)

\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{\sqrt{3}}{2}\) có GTNN là \(\frac{\sqrt{3}}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

Vậy \(P_{min}=\frac{\sqrt{3}}{2}\) tại \(x=y=z=\frac{1}{\sqrt{3}}\)

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

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

\(=\left(x-y\right)^2\left(z^2-1\right)+\left(y-z\right)^2\left(x^2-1\right)+\left(z-x\right)^2\left(y^2-1\right)\le0\)

\(\text{( Do }x^2;y^2;z^2\le1\text{)}\)

\(\Rightarrow2P\le2\Rightarrow P\le1\)

\(\text{Dấu bằng xảy ra khi và chỉ khi 1 trong 3 số bằng 1; 2 số còn lại bằng 0.}\)