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\(a,\left(x+y+z\right)^3-x^3-y^3-z^3\\ =\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\\ =\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =\left(x+y\right)\left(3xy+3xz+3yz+3z^2\right)\\ =3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\\ =3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(b,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\\ =0\left(x^2+y^2+z^2-xz-yz-xy\right)=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)
Ta có:\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)=0\)
\(\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\right]=0\)
\(x+y+z=0\)hoặc \(x=y=z\)(Đpcm)
Ta có: \(\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)
\(=\frac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz}{x+y+z}\)
\(=\frac{\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)}{x+y+z}\)
\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-zx-3xy\right)}{x+y+z}\)
\(=x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\left(\forall x,y,z\right)\)
=> đpcm
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
Theo đề bài ta có:
\(\left\{\begin{matrix}x\ge xy\\y\ge yz\\z\ge xz\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x-xy\ge0\\y-yz\ge0\\z-xz\ge0\end{matrix}\right.\)
\(\Rightarrow x+y+z-xy-yz-xz\ge0\)
Xét tích
\(\left(1-x\right)\left(1-y\right)\left(1-z\right)=-\left(x+y+z-xy-yz-xz-1+xyz\right)\ge0\)
\(\Rightarrow x+y+z-xy-yz-xz\le1-xyz\)
\(0\le xyz\le1\) nên \(1-xyz\le1\)
Vậy \(x+y+z-xy-yz-xz\le1\)
ta có x+y+z=0
=> x+y=-z
=> (x+y)^3=(-z)^3
=> x^3+y^3+3xy(x+y)=-z^3
x^3+y^3+z^3+3xy(x+y)=0
x^3+y^3+z^3-3xyz=0
=> x^3+y^3+z^3=3xyz
\(VT-VP=\frac{x^8+y^8+z^8}{x^3y^3z^3}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\equiv\frac{F\left(x;y;z\right)}{x^3y^3z^3}\)
Ta có: \(F\left(x;y;z\right)=G\left(x;y;z\right)+M\left(x;y;z\right)\ge0\)
Với \(G\left(x;y;z\right)=\)
(hiển nhiên không âm)
\(M\left(x;y;z\right)=\)
(cũng hiển nhiên không âm)
Ta có đpcm.
x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz
= (x+y)^3 + z^3 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z)
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy]
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
=1/2(x+y+z)(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2)
=1/2(x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2]
mà x^3 + y^3 + z^3 - 3xyz=0
<=> x+y+z=0
Vậy ...
Chúc bạn học tốt .
hoặc (x-y)^2+(y-z)^2+(x-z)^2 =0 mà (x-y)^2,(y-z)^2,(x-z)^2 >=0 mọi x,y,z
=> x-y=y-z=x-z=0 => x=y=z