\(x^2+y^2=1+xy\Rightarrow x^2+y^2-xy=1\)

Ta có: \(1+xy=x^2+y^2\ge2xy\Rightarrow xy\le1\)

\(1+xy=x^2+y^2\ge-2xy\Rightarrow xy\ge-\dfrac{1}{3}\)

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

\(=x^2+y^2+xy-2x^2y^2=-2x^2y^2+2xy+1\)

Đặt \(a=xy\Rightarrow P=f\left(a\right)=-2a^2+2a+1\)

Xét hàm \(f\left(a\right)=-2a^2+2a+1\) trên \(\left[-\dfrac{1}{3};1\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-\dfrac{1}{3};1\right]\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow M=\dfrac{3}{2}\) ; \(m=\dfrac{1}{9}\) \(\Rightarrow Mm=\dfrac{1}{6}\)

Đặt A=\(\left|2x-3y\right|+\left|4z-3x\right|+\left|xy+yz+xz-2484\right|\)

Ta có \(\left|2x-3y\right|\ge0;\left|4z-3x\right|\ge0;\left|xy+yz+xy-2484\right|\ge0\)

\(\Rightarrow A\ge0\Rightarrow Amin=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-3y=0\\4z-3x=0\\xy+yz+xz-2484=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{12}=\frac{y}{8}\left(1\right)\\\frac{x}{4}=\frac{z}{3}\Rightarrow\frac{x}{12}=\frac{z}{9}\left(2\right)\\xy+yz+xz=2484\left(3\right)\end{cases}}}\)

Từ (1)(2)\(\Rightarrow\frac{x}{12}=\frac{y}{8}=\frac{z}{9}=k\left(k\ne0\right)\)

\(\Rightarrow x=12k;y=8k;z=9k\)

Thay vào 3 ta có \(12.8.k^2+8.9.k^2+12.9.k^2=2484\)

\(\Rightarrow k^2\left(12.8+8.9+12.9\right)=2484\)

\(\Rightarrow k^2.276=2484\)

\(\Rightarrow k^2=9=\left(\pm3\right)^2\)

\(\Rightarrow k=\pm3\)

+Nếu k =3 thì      x=36          ;                  y=24                        ;                      z=27

+Nếu k = -3thì    x=-36          ;                   y=-24                      ;                        z=-27

Vậy \(Amin=0\Leftrightarrow\left(x;y;z\right)\in\left\{\left(36;24;27\right);\left(-36;-24;-27\right)\right\}\)

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

Cách giải khác:

Dư đoán khi \(x=y=z=\frac{1}{\sqrt{3}}\) thì ta được \(P_{Min}=1\)

Thật vậy cần chứng minh \(Σ\frac{1}{4x^2-yz+2}\ge1\LeftrightarrowΣ\left(\frac{1}{4x^2-yz+2}-\frac{1}{3}\right)\ge0\)

\(\LeftrightarrowΣ\frac{1-4x^2+yz}{4x^2-yz+2}\ge0\LeftrightarrowΣ\frac{xy+xz+2yz-4x^2}{4x^2-yz+2}\ge0\)

\(\LeftrightarrowΣ\frac{\left(z-x\right)\left(2x+y\right)-\left(x-y\right)\left(2x+z\right)}{4x^2-yz+2}\ge0\)

\(\LeftrightarrowΣ\left(x-y\right)\left(\frac{2y+z}{4y^2-xz+2}-\frac{2x+z}{4x^2-yz+2}\right)\ge0\)

\(\LeftrightarrowΣ\left(x-y\right)^2\left(z^2+2xy+2\right)\left(z^2-xy+2\right)\ge0\)

3/2 nha

Let \(\hept{\begin{cases}xy=a\\yz=b\\xz=c\end{cases}}\Rightarrow a+b+c=1\)

\(\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{a}\)

Hence \(S=Σ\frac{1}{\frac{4ac}{b}-b+2}=Σ\frac{1}{\frac{4ac}{b}-b+2\left(a+b+c\right)}\)

\(=Σ\frac{1}{\frac{4ac}{b}+2a+b+2c}=Σ\frac{b}{4ac+2ab+b^2+2bc}\)

\(=Σ\frac{b}{\left(2a+b\right)\left(2c+b\right)}\geΣ\frac{4b}{\left(2a+2b+2c\right)^2}=Σ\frac{b}{\left(a+b+c\right)^2}=1\)

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

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

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

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 \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=1+2\left(ab+bc+ca\right).\)

\(\Rightarrow A=\left(ab+bc+ca\right)=\frac{1}{2}\left(a+b+c\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)với mọi a,b,c

Vậy A nhỏ nhất bằng -1/2 khi a+b+c =0

Ta có : \((x-\dfrac{1}{3})^2+(y-\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2>=0\)

\(=>x^2+y^2+z^2-\dfrac{2}{3}(x+y+z)+\dfrac{1}{3}\ge0\)

\(=>x^2+y^2+z^2+\dfrac{1}{3}\ge\dfrac{2}{3}(x+y+z)\)

\(=>1+\dfrac{1}{3}=\dfrac{4}{3}\ge\dfrac{2}{3}(x+y+z)\)

\(=>x+y+z\le2\)

Do đó : \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca).\)

\(=>A=(ab+ac+bc)=\dfrac{1}{2}(a+b+c)^2-\dfrac{1}{2}\le\dfrac{1}{2}.2^2-\dfrac{1}{2}=\dfrac{3}{2}\)

Ta có : \((x-\dfrac{1}{\sqrt{3}})^2+(y-\dfrac{1}{\sqrt{3}})^2+(z-\dfrac{1}{\sqrt{3}})^2>=0\)

\(=>x^2+y^2+z^2-\dfrac{2}{\sqrt{3}}(x+y+z)+1\ge0\)

\(=>x^2+y^2+z^2+1\ge\dfrac{2}{\sqrt{3}}(x+y+z)\)

\(=>1+1=2\ge\dfrac{2}{\sqrt{3}}(x+y+z)\)

\(=>x+y+z\le\sqrt{3}\)

Do đó : \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca).\)

\(=>A=(ab+ac+bc)=\dfrac{1}{2}(a+b+c)^2-\dfrac{1}{2}\le\dfrac{1}{2}.\sqrt{3}^2-\dfrac{1}{2}=\dfrac{2}{2}=1\)