1. 1/x + 2/1-x = (1/x - 1) + (2/1-x - 2) + 3

= 1-x/x + (2-2(1-x))/1-x  + 3

= 1-x/x + 2x/1-x + 3    >= 2√2 + 3

Dấu "=" xảy ra khi x =√2 - 1

2. a = √z-1, b = √x-2, c = √y-3 (a,b,c >=0)

=> P = √z-1 / z + √x-2 / x + √y-3 / y 

= a/a^2+1 + b/b^2+2 + c/c^2+3

a^2+1 >= 2a              => a/a^2+1 <= 1/2

b^2+2 >= 2√2 b          => b/b^2+2 <= 1/2√2

c^2+3 >= 2√3 c            => c/c^2+3 <= 1/2√3

=> P <= 1/2 + 1/2√2 + 1/2√3

Dấu = xảy ra khi a^2 = 1, b^2 = 2, c^2 =3

<=> z-1 = 1, x-2 = 2, y-3 = 3

<=> x=4, y=6, z=2

x=6

y=8

z=10

Xin lỗi bạn vì mình không biết cách để tính theo cách tích ở tử.

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{xy}{12}=\dfrac{yz}{20}=\dfrac{zx}{15}=\dfrac{xy+yz+zx}{12+20+15}=\dfrac{188}{47}=4\)

\(\Rightarrow\left\{{}\begin{matrix}xy=4.12\\yz=4.20\\zx=4.15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}xy=48\\yz=80\\zx=60\end{matrix}\right.\)

\(\Rightarrow x^2.y^2.z^2=48.80.60\)

\(\Rightarrow\left(xyz\right)^2=480^2\)

\(\Rightarrow xyz=480\)

\(\Rightarrow\left\{{}\begin{matrix}x=480:80\\y=480:60\\z=480:48\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\\z=10\end{matrix}\right.\)

Vậy...

x + y + z = 6 => (x + y + z)² = 36

=> x² + y² + z² + 2(xy + yz + zx) = 36

=> x² + y² + z² = 36 - 2.12 = 12

=> x² + y² + z² = xy + yz + zx

Ta có VT \(\ge\) VP. Dấu "=" xảy ra <=> x = y = z

Thay vào hệ ta có (x; y; z) = (2; 2; 2)

\(P=\dfrac{1}{x^2+y^2+z^2}+\dfrac{2023}{xy+yz+zx}\)

\(=\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{2021}{xy+yz+zx}\)

\(\ge\dfrac{9}{\left(x+y+z\right)^2}+\dfrac{2021}{\dfrac{\left(x+y+z\right)^2}{3}}\)\(=9+\dfrac{2021}{\dfrac{1}{3}}=6072\)

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

Ta có:

+) \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\left(\text{Cô si}\right)\)

+) \(\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}\)

\(\ge\dfrac{9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=\dfrac{9}{\left(x+y+z\right)^2}\left(\text{Svácxơ}\right)\)

Chứng minh gì bạn?

\(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=\dfrac{12}{5}\\\dfrac{yz}{y+z}=\dfrac{18}{5}\\\dfrac{zx}{z+x}=\dfrac{36}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{5}{12}\\\dfrac{y+z}{yz}=\dfrac{5}{18}\\\dfrac{z+x}{zx}=\dfrac{13}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{12}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{18}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{13}{36}\end{matrix}\right.\)

Cộng vế theo vế ta thu được :

\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{19}{18}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{19}{36}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{4}\\\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{z}=\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=9\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(4;6;9\right)\)

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

Áp dụng BĐT AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

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

Áp dụng bất đẳng thức AM - GM và kết hợp với giả thiết x + y + z = 3 ta có:

\(B=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}+\sqrt{\dfrac{yz}{yz+x\left(x+y+z\right)}}+\sqrt{\dfrac{zx}{zx+y\left(x+y+z\right)}}\)

\(B=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\dfrac{yz}{\left(y+x\right)\left(z+x\right)}}+\sqrt{\dfrac{zx}{\left(z+y\right)\left(z+x\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}+\dfrac{y}{y+x}+\dfrac{z}{z+x}+\dfrac{z}{z+y}+\dfrac{x}{z+x}\right)\)

\(B\le\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

Vậy...

Đề bài sai, biểu thức này ko có min