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Cách 1:
Ta có \(A=xy+yz+2zx\)
\(\Rightarrow A+1=x^2+y^2+z^2+xy+yz+2zx\)
\(=\left(x+z+\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge0\)
\(\Rightarrow A\ge-1\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}y=0\\x=-z\end{cases}}\)
Ta có : \(\left(x+y+z\right)^2\ge0\)
\(\Rightarrow xy+yz+zx\ge\frac{-\left(x^2+y^2+z^2\right)}{2}=-\frac{1}{2}\)
Lại có : \(\left(x+z\right)^2\ge0\Rightarrow xz\ge\frac{-\left(x^2+z^2\right)}{2}=\frac{y^2-1}{2}\ge-\frac{1}{2}\)
Khi đó : \(xy+yz+2zx\ge-1\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=o\\x^2=z^2=\frac{1}{2}\end{cases}}\)
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ta có \(xy\le\left(\frac{x+y}{2}\right)^2\) và \(yz+xz=z\left(x+y\right)\le\frac{z^2+\left(x+y\right)^2}{2}\)
\(\Rightarrow5=xy+yz+xz\le\left(\frac{x+y}{2}\right)^2+\frac{z^2+\left(x+y\right)^2}{2}=\frac{3}{4}\left(x+y\right)^2+\frac{1}{2}z^2\)
Xét \(3x^2+3y^2+z^2\ge\frac{3}{2}\left(x+y\right)^2+z^2=2\left(\frac{3}{4}\left(x+y\right)^2+\frac{1}{2}z^2\right)\ge2\cdot5=10\)
dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\z=x+y\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=\pm1\\z=\pm2\end{cases}}}\)
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Cauchy-Schwarz : \(\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)\ge\left(xy+yz+zx\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2\ge\left|xy+yz+zx\right|\ge xy+yz+zx\)(1)
Mặt khác :
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2+y^2+z^2=9-2\left(xy+yz+zx\right)\)
Kết hợp (1)
=> \(9-2\left(xy+yz+xz\right)\ge xy+yz+zx\)
\(\Leftrightarrow3\left(xy+yz+zx\right)\le9\)
\(\Leftrightarrow xy+yz+zx\le3\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\\x+y+z=3\end{cases}}\)<=> x=y=z=1
Vậy MaxM=3 khi x=y=z=1
Ta có :
\(M=x^4+y^4+z^4=\left(x^4+\frac{1}{9}\right)+\left(y^4+\frac{1}{9}\right)+\left(z^4+\frac{1}{9}\right)-\frac{1}{3}\)
Áp dụng BĐT \(a^2+b^2\ge2ab\) ( "=" khi a=b ) , ta có :
\(M\ge\frac{2}{3}x^2+\frac{2}{3}y^2+\frac{2}{3}z^2-\frac{1}{3}\)
\(\Rightarrow M\ge\frac{1}{3}\left(2x^2+2y^2+2z^2\right)-\frac{1}{3}\)
\(\Rightarrow M\ge\frac{1}{3}\left[\left(x^2+y^2\right)+\left(y^2+z^2\right)+\left(x^2+z^2\right)\right]-\frac{1}{3}\)
\(\Rightarrow M\ge\frac{2}{3}.\left(xy+yz+xz\right)-\frac{1}{3}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) ( Vì xy+yz+xz=1 )
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Vậy \(GTNN_M=\frac{1}{3}\) khi \(x=y=z=\frac{1}{\sqrt{3}}\)
( Ko bít đúng Ko ) :)
cảm ơn nha