\(A=\frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x}\)

Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(6=x+y+z+xy+yz+zx\le x+y+z+\frac{\left(x+y+z\right)^2}{3}\)

\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)

\(\Leftrightarrow\left(x+y+z-3\right)\left(x+y+z+6\right)\ge0\)

\(\Leftrightarrow x+y+z\ge3\)(vì \(x,y,z>0\)) 

Ta có: \(\frac{x^3}{y}+y+1\ge3x,\frac{y^3}{z}+z+1\ge3y,\frac{z^3}{x}+x+1\ge3z\)

Suy ra \(A\ge2\left(x+y+z\right)-3\ge2.3-3=3\)

Dấu \(=\)xảy ra khi \(x=y=z=1\).

Ta có: \(\left|2x+3y\right|\ge0\)\(\forall x,y\inℝ\); \(\left|4y+5z\right|\ge0\)\(\forall y,z\inℝ\); \(\left|xy+yz+zx+110\right|\ge0\)\(\forall x,y,z\inℝ\)

Nên: \(P=\left|2x+3y\right|+\left|4y+5z\right|+\left|xy+yz+xz+110\right|\ge0\)\(\forall x,y,z\inℝ\)

Dấu " = " xảy ra <=> \(\left|2x+3y\right|+\left|4y+5z\right|+\left|xy+yz+xz+110\right|=0\)

Có: \( \left|2x+3y\right|=0\)\(\Leftrightarrow2x+3y=0\)\(\Leftrightarrow2x=-3y\)\(\Leftrightarrow\frac{x}{-3}=\frac{y}{2}\)

\(\left|4y+5z\right|=0\)\(\Leftrightarrow4y+5z=0\)\(\Leftrightarrow4y=-5z\)\(\Leftrightarrow\frac{y}{-5}=\frac{z}{4}\)

\(\left|xy+yz+zx+110\right|=0\)\(\Leftrightarrow xy+yz+zx+110=0\)\(\Leftrightarrow xy+yz+zx=-110\)

Lại có: \(\frac{x}{-3}=\frac{y}{2}\)\(\Rightarrow\frac{x}{15}=\frac{y}{-10}\) (1) ;  \(\frac{y}{-5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{-10}=\frac{z}{8}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{-10}=\frac{z}{8}=k\)=> x = 15k ; y = (-10) . k ; z = 8k

Ta có: \(xy+yz+zx=-110\)\(\Rightarrow15k\left(-10\right)k+8k\left(-10\right)k+8k.15k=-110\)

\(\Rightarrow k^2\left(-150\right)+k^2\left(-80\right)+120k^2=-110\)

\(\Rightarrow k^2\left(-110\right)=-110\)\(\Rightarrow k^2=1\)\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

+) Th1: k = 1   

Có: x = 15k = 15 . 1 = 15

y = (-10) . k = (-10) . 1 = -10

z = 8k = 8 . 1 = 8

+) Th2: k = -1

Có: x = 15k = 15 . (-1) = -15 

y = (-10) . k = (-10) . (-1) = 10

z = 8k = 8 . (-1) = -8

Vậy GTNN P = 0 <=> (x; y; z) = (15; -10; 8) hoặc (x; y; z) = (-15; 10; -8)

 với mọi x, y, z ta có: 
(x-y)^2 +(y-z)^2+ (z-x)^2>=0 
<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0 
<=>x^2 + y^2 +z^2 - xy -yz -zx >=0 
<=>(x+y+z)^2 >= 3(x+y+z) 
<=>[(x+y+z)^2]/3 >= xy+yz+ zx 
=>xy +yz + zx <=3 
dấu = xảy ra khi x=y=z =1

tk nha bạn

thank you bạn

(^_^)

le anh tu giỏi quá, làm đúng rồi

Bạn Hồ Thị Hà Giang làm theo cách của bạn ấy nha

Ai thấy mình nói đúng thì nha

với mọi x, y, z ta có: 
(x-y)^2 +(y-z)^2+ (z-x)^2>=0 
<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0 
<=>x^2 + y^2 +z^2 - xy -yz -zx >=0 
<=>(x+y+z)^2 >= 3(x+y+z) 
<=>[(x+y+z)^2]/3 >= xy+yz+ zx 
=>xy +yz + zx <=3 
dấu = xảy ra khi x=y=z =1

>_<

CM được BĐT : \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\)\(\Rightarrow\frac{yz+xy+xz}{xyz}\ge9\)

\(\Rightarrow xy+yz+xz-9xyz\ge0\)

\(\Rightarrow A\ge-3xyz\ge3.\left[-\left(\frac{x+y+z}{3}\right)^3\right]=3.\left(-\frac{1}{27}\right)=\frac{-1}{9}\)

Vậy GTNN của A là \(\frac{-1}{9}\)khi \(x=y=z=\frac{1}{3}\)