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\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=a^2+b\)
\(\Rightarrow xy+yz+xz=\dfrac{a^2+b}{2}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\Rightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{c}\)
\(\Rightarrow xyz=c\left(xy+yz+xz\right)\)
\(\Rightarrow xyz=\dfrac{\left(a^2+b\right)c}{2}\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=a\left(b-\dfrac{a^2+b}{2}\right)+3\dfrac{\left(a^2+b\right)c}{2}\)
\(\Rightarrow x^3+y^3+z^3=a\dfrac{\left(b-a^2\right)}{2}+3\dfrac{\left(a^2+b\right)c}{2}\)
Do \(xyz\ne0\) ta có:
\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)
Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)
Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
Ta có:\(xy\le\dfrac{\left(x+y\right)^2}{4}\)(tự cm)
Áp dụng vào \(\Rightarrow P=\dfrac{xy}{z+1}+\dfrac{yz}{x+1}+\dfrac{zx}{y+1}\)
\(P\le\dfrac{\left(x+y\right)^2}{4z+4}+\dfrac{\left(y+z\right)^2}{4x+4}+\dfrac{\left(z+x\right)^2}{4y+4}\le\dfrac{\left[2\left(x+y+z\right)\right]^2}{4\left(x+y+z+3\right)}=\dfrac{2^2}{4\cdot4}=\dfrac{1}{4}\)
\(\Rightarrow MAXP=\dfrac{1}{4}\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Lời giải:
Từ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
$\Rightarrow xy+yz+xz=0$
Khi đó:
$x^2+2yz=x^2+yz-xz-xy=(x^2-xy)-(xz-yz)=x(x-y)-z(x-y)=(x-z)(x-y)$
Tương tự với $y^2+2zx, z^2+2xy$ thì:
$P=\frac{yz}{(x-z)(x-y)}+\frac{xz}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$
$=\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\frac{-[yz(y-z)+xz(z-x)+xy(x-y)]}{-[xy(x-y)+yz(y-z)+xz(z-x)]}=1$
Lời giải:
Cách 1:
Áp dụng BĐT S.Vacxo ta có:
\(\frac{1}{xy+1}+\frac{1}{1+yz}+\frac{1}{1+xz}\geq \frac{9}{1+xy+1+yz+1+xz}=\frac{9}{3+xy+yz+xz}(1)\)
Theo BĐT Cauchy ta có bổ đề quen thuộc:
\(xy+yz+xz\leq x^2+y^2+z^2\leq 3(2)\)
Từ \((1);(2)\Rightarrow \frac{1}{xy+1}+\frac{1}{yz+1}+\frac{1}{xz+1}\geq \frac{9}{3+xy+yz+xz}\geq \frac{9}{3+3}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow x=y=z=1\)
Cách 2:
Áp dụng BĐT Cauchy cho các số dương:
\(\frac{1}{xy+1}+\frac{xy+1}{4}\geq 2.\sqrt{\frac{1}{xy+1}.\frac{xy+1}{4}}=1\)
\(\frac{1}{yz+1}+\frac{yz+1}{4}\geq 2.\sqrt{\frac{1}{yz+1}.\frac{yz+1}{4}}=1\)
\(\frac{1}{xz+1}+\frac{xz+1}{4}\geq 2.\sqrt{\frac{1}{xz+1}.\frac{xz+1}{4}}=1\)
Cộng tất cả các BĐT trên theo vế và rút gọn:
\(\Rightarrow \frac{1}{xy+1}+\frac{1}{yz+1}+\frac{1}{xz+1}\geq \frac{9-(xy+yz+xz)}{4}\geq \frac{9-3}{4}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)