Cho 3 số x,y,z thoả mãn : x+y+z=3 . Tính Max của xy + yz + xz ?
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Áp dụng BĐT Cauchy Schwarz ta có:
\(\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)\ge\left(xy+yz+xz\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2\ge\left|xy+yz+xz\right|\ge xy+yz+xz\left(1\right)\)
Mặt khác:
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2=9-2\left(xy+yz+xz\right)\)
Kết hợp với \(\left(1\right)\Rightarrow9-2\left(xy+yz+xz\right)\ge xy+yz+xz\)
\(\Leftrightarrow3\left(xy+yz+xz\right)\le9\Leftrightarrow xy+yz+xz\le3\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\\x+y+z=3\end{cases}}\Leftrightarrow x=y=z=1\)
Vậy \(Max\) biểu thức là \(3\Leftrightarrow x=y=z=1\)
Với \(x,y,z\)ta có :
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2>=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge=0\)
\(x^2+y^2+z^2-xy-yz-zx\ge=0\)
\(\left(y+x+z\right)^2\ge=3\left(x+y+z\right)\)
\(\frac{\left[\left(x+y+z\right)^2\right]}{3}\ge=xy+zx+yz\)
\(\Rightarrow xy+yz+zx\le=3\)
Dấu \(=\)xảy ra khi \(x=y=z=1\)
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Với a,b,c dưog thì \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}>=\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
\(P>=\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+\sqrt{1+x^3}+\sqrt{1+y^3}+\sqrt{1+z^3}}\)
\(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+x^2\right)}< =\dfrac{2+x^2}{2}\)
Dấu = xảy ra khi x=2
=>\(P>=\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2+6}=\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2+6}\)
Đặt t=(x+y+z)^2(t>=36)
=>P>=2t/t-6
Xét hàm số \(f\left(t\right)=\dfrac{t}{t+6}\left(t>=36\right)\)
\(f'\left(t\right)=\dfrac{6}{\left(t+6\right)^2}>=0,\forall t>=36\)
=>f(t) đồng biến
=>f(t)>=f(36)=6/7
=>P>=12/7
Dấu = xảy ra khi x=y=z=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Tá có : $(x+1).(y+1).(z+1) = (x-1).(y-1).(z-1)$
$\to xyz+1+x+y+z+xy+yz+zx =xyz + x + y + z -xy-yz-zx-1$
$\to 2.(xy+yz+zx) = -2$
$\to xy+yz+zx=-1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(P=xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=3\).
Đẳng thức xảy ra khi x = y = z = 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{3}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{x^2+y^2+z^2+2xy+yz+xz}=\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{3}\right)^2\)
(*) ta CM :\(\left(\sqrt{6}+\sqrt{3}\right)^2>14\)
TA có \(\left(\sqrt{6}+\sqrt{3}\right)^{^2}=6+3+2\sqrt{18}=9+6\sqrt{2}>9+5=14\)
=> \(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}>14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Rightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\ge0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)
\(\Rightarrow2\left(x^2+y^2+z^2-xy-yz-xz\right)\ge0\)
\(\Rightarrow x^2+y^2+z^2-xy-yz-xz\ge0\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz-3xy-3yz-3xz\ge0\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\)
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\)
\(\Rightarrow xy+yz+xz\le\frac{3^2}{3}=3\)
=> \(P_{min}=xy+yz+xz=3\Leftrightarrow x=y=z=1\)
Vậy ...................
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Cho 3 số x,y,z thoả mãn : x+y+z=3 . Tính Max của xy + yz + xz ? < giá trị lớn nhất>? | Yahoo Hỏi & Đáp
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