Question

$\text{Cho $x,y,z$ không âm thỏa mãn:}\\\begin{cases}x,y,x\le1\\x+y+z=\dfrac32\end{cases} \ \text{Tìm $Max$:}\\P=x^2+y^2+z^2$

Answer 1

`0<=y,z<=1`

`=>1-y,1-z>=0`

`=>(1-y)(1-z)>=0`

`=>1-y-z+yz>=0`

`=>yz>=y+z-1`

`=>2yz>=2x+2z-2`

`=>P=x^2+y^2+z^2`

`=>P=x^2+(y^2+2yz+z^2)-2yz`

`=>P=x^2+(y+z)^2-2yz`

`=>P<=x^2-2(y+z-1)+(3/2-x)^2`

`=>P<=(3/2-x)^2-2(1/2-x)+x^2`

`=>P<=9/4-3x+x^2-1+2x+x^2`

`=>P<=5/4+2x^2-x`

Giả sử:

`x<=y<=z`

`=>x+x+x<=x+y+z=3/2`

`=>3x<=3/2`

`=>x<=1/2`

`0<=x<=1/2=>2x^2-x<=0`

`=>P<=5/4`

Dấu "=" xảy ra khi `(x,y,z)=(0,1,1/2)` và các hoán vị

Answer 2

Ta có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x,y,z\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2+x^2+y^2+z^2\ge x^2+y^2+z^2+2xy+2yz+2xz\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\ge\dfrac{9}{4}:3=\dfrac{9}{4}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)

Dấu '=' xảy ra khi \(x=y=z=\dfrac{1}{4}\)

Vậy: \(P_{max}=\dfrac{3}{4}\) khi \(x=y=z=\dfrac{1}{4}\)

Answer 3

`0<=y,z<=1`

`=>1-y,1-z>=0`

`=>(1-y)(1-z)>=0`

`=>1-y-z+yz>=0`

`=>yz>=y+z-1`

`=>2yz>=2x+2z-2`

`=>P=x^2+y^2+z^2`

`=>P=x^2+(y^2+2yz+z^2)-2yz`

`=>P=x^2+(y+z)^2-2yz`

`=>P<=x^2-2(y+z-1)+(3/2-x)^2`

`=>P<=(3/2-x)^2-2(1/2-x)+x^2`

`=>P<=9/4-3x+x^2-1+2x+x^2`

`=>P<=5/4+2x^2-x`

Giả sử:

`x<=y<=z`

`=>x+x+x<=x+y+z=3/2`

`=>3x<=3/2`

`=>x<=1/2`

`0<=x<=1/2=>2x^2-x<=0`

`=>P<=5/4`

Dấu "=" xảy ra khi `(x,y,z)=(0,1,1/2)` và các hoán vị