\(xy+yz+zx=xyz\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\) thì

\(\hept{\begin{cases}a+b+c=1\\P=\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{1}{16}\end{cases}}\)

Ta co:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{64}+\frac{1+c}{64}\ge\frac{3a}{16}\)

\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{3a}{16}-\frac{b}{64}-\frac{c}{64}-\frac{1}{32}\)

Từ đây ta co:

\(P\ge\left(a+b+c\right)\left(\frac{3}{16}-\frac{1}{64}-\frac{1}{64}\right)-\frac{3}{32}=\frac{1}{16}\)

khó quá nguyen van hung

Ta chứng minh \(\frac{x^4+y^4}{x^2+y^2}\ge\frac{\frac{\left(x^2+y^2\right)^2}{2}}{x^2+y^2}=\frac{x^2+y^2}{2}\)

Tương tự và cộng lại

\(\Rightarrow VT\ge x^2+y^2+z^2\ge xy+xz+yz=3\)

chứng minh kiểu j vậy bạn ? , Chỉ mình rõ hơn được không ? 

Giả sử z = min{x,y,z} \(\Rightarrow4=x+y+z+xyz\ge z^3+3z\Leftrightarrow\left(z-1\right)\left(z^2+z+4\right)\le0\Rightarrow z\le1\)(*)

Chọn t thỏa mãn \(\hept{\begin{cases}x+y+z+xyz=2t+z+t^2z\\2t+z+t^2z=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y-2t=\left(t^2-xy\right)z\left(1\right)\\2t+z+t^2z=4\left(2\right)\end{cases}}\)

Giả sử \(t^2< xy\Rightarrow2t>x+y\ge2\sqrt{xy}\Rightarrow t^2>xy\) (mâu thuẫn với giả sử)

Vậy \(t^2\ge xy\Rightarrow x+y\ge2t\). Đặt  P = f(a;b;c). Xét hiệu:

\(f\left(x;y;z\right)-f\left(t;t;z\right)=z\left(x+y-2t\right)-\left(t^2-xy\right)\)

\(=z^2\left(t^2-xy\right)-\left(t^2-xy\right)=\left(z^2-1\right)\left(t^2-xy\right)\le0\)

Vậy: \(P=f\left(x;y;z\right)\le f\left(t;t;z\right)=t^2+2tz\)

 Từ \(\left(2\right)\Rightarrow z=\frac{\left(4-2t\right)}{t^2+1}.\text{Do }z\ge0\Rightarrow4-2t\ge0\Rightarrow t\le2\)

Mặc khác do (*): \(\Rightarrow4=2t+z+t^2z\le t^2+2t+1\Rightarrow\left(t+3\right)\left(t-1\right)\ge0\Rightarrow2\ge t\ge1\)

Vậy ta tìm max của: \(f\left(t;t;z\right)=f\left(t;t;\frac{4-2t}{t^2+1}\right)=t^2+\frac{2t\left(4-2t\right)}{t^2+1}\)

Dễ thấy hàm số này đồng biến suy ra \(f\left(t;t;\frac{4-2t}{t^2+1}\right)\) đạt max khi t = 2. Khi đó \(P=f\left(a;b;c\right)\le f\left(t;t;\frac{4-2t}{t^2+1}\right)\le4\)

Đẳng thức xảy ra khi \(\left(x;y;z\right)=\left(2;2;0\right)\) và các hoán vị.

P/s: em hết cách rồi nên đành chơi kiểu này:(

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

\(=\left(3-x\right)^2+\left(3-y\right)^2+\left(3-z\right)^2\)

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Dễ dàng CM được \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=3\)

=>\(2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge12\)

=> dpcm

Ta có: \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\)

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

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

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

Mà \(x+y+z=3\Rightarrow\hept{\begin{cases}x+y=3-z\\y+z=3-x\\x+z=3-y\end{cases}}\)

\(\Rightarrow\left(1\right)=\left(3-z\right)^2+\left(3-x\right)^2+\left(3-y\right)^2\)

\(=9-6z+z^2+9-6x+x^2+9-6y+y^2\)

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Áp dụng BĐT Cauchy cho 3 số:

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

\(\Rightarrow9+x^2+y^2+z^2\ge12\)

hay \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\ge12\)

\(\Leftrightarrow x^2+y^2+z^2+xy+yz+xz\ge6\left(đpcm\right)\)

Sử dụng bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^2\ge3\left(\frac{xy}{z}.\frac{yz}{x}+\frac{yz}{x}.\frac{zx}{y}+\frac{zx}{y}.\frac{xy}{z}\right)=3\left(x^2+y^2+z^2\right)=3\)

\(\Rightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge\sqrt{3}\)

Áp dụng BĐT Cauchy-Schwarz ta có: 

\(T=\frac{1}{xy}+\frac{1}{xz}\ge\frac{\left(1+1\right)^2}{xy+xz}=\frac{4}{xy+xz}\)

Từ \(x+y+z=3\Rightarrow y+z=4-x\)

\(\Rightarrow T\ge\frac{4}{xy+xz}=\frac{4}{x\left(y+z\right)}=\frac{4}{x\left(4-x\right)}=\frac{4}{-x^2+4x}\)

Lại có: \(-x^2+4x=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\le4\)

\(\Rightarrow T\ge\frac{4}{-x^2+4x}\ge\frac{4}{4}=1\)

Đẳng thức xảy ra khi \(x=2;y=z=1\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{4}{x\left(y+z\right)}\ge\frac{4}{\frac{\left(x+y+z\right)^2}{4}}=1\).