Đặt \(\left(x,y,z\right)=\left(a+1,b+1,c+1\right)\Rightarrow a,b,c\ge0\)

Ta có : 

\(3x^2+4y^2+5z^2=52\Leftrightarrow3\left(a+1\right)^2+4\left(b+1\right)^2+5\left(c+1\right)^2=52\)

\(\Leftrightarrow3a^2+4b^2+5c^2+6a+8b+10c=40\)

\(\Leftrightarrow5\left(a+b+c\right)^2+10\left(a+b+c\right)=40+2a^2+b^2+10\left(ab+bc+ac\right)+4a+2b\)

\(\Rightarrow5\left(a+b+c\right)^2+10\left(a+b+c\right)\ge40\Leftrightarrow a+b+c\ge2\)

Do đó \(x+y+z=a+b+c+3\ge5\)

Vậy \(F_{min}=5\Leftrightarrow x=y=1;z=3\)

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Cho các số thực x y z ge1 thỏa mãn 3x 2 4y 2 5z 2 52 Tìm ...

Tìm GTNN của F=x+y+z biết 3x^2+4y^2+5z^2-52 - H7.net

Search mạng đầy vler :333

vì x,y \(\ge\)1 nên ( x - 1 )( y - 1 ) \(\ge\)0 \(\Leftrightarrow\)xy \(\ge\)x + y - 1

Tương tự : yz \(\ge\)y + z - 1 ; xz \(\ge\)x + z - 1

Cộng lại, ta được : xy + yz + xz \(\ge\)2 ( x + y + z ) - 3

Do đó : 5 ( x + y + z )² = 5 ( x² + y² + z² ) + 10 ( xy + yz + xz ) \(\ge\)52 + 2x² + y² + 10 . [ 2( x+y+z ) - 3 ]

\(\ge\)52 + 2 + 1 + 20 ( x+y+z ) - 30 = 25 + 20 ( x+ y + z )

\(\Rightarrow5\left(x+y+z\right)^2-20\left(x+y+z\right)-25\ge0\)

\(\Rightarrow x+y+z\ge5\). Dấu " = " xảy ra khi x = y = 1 ; z = 3

Từ giả thiết ta có :

\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

ta có : \(Q=\frac{y+2}{x^2}+\frac{z+2}{y^2}+\frac{x+2}{z^2}\)

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

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

\(\ge\frac{2\left(x+1\right)}{zx}+\frac{2\left(y+1\right)}{xy}+\frac{2\left(z+1\right)}{yz}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2\)

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

Dấu " = " xảy ra khi và chỉ khi a = b = c

Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=3\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\sqrt{3}\)

Do đó : \(Q\ge\sqrt{3}+2\). Dấu " = " xảy ra 

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\\z+y+z=xyz\end{cases}\Leftrightarrow x=y=z=\sqrt{3}}\)

Vậy Min \(Q=\sqrt{3}+2\)khi \(x=y=z=\sqrt{3}\)

 đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)

\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)

BBDT AM-GM 

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)

theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)

vì \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)

\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)

dấu"=" xảy ra<=>x=y=z=1/3

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 \(x=y=z=\frac{1}{\sqrt{3}}\)

đề đúng ko v

đúng đó bạn ạ

úi lộn k phải 3² mà là 3x^2