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Ta có :
\(x^3+x^2z+y^2z-xyz+y^3\)
\(=x^3+y^3+x^2z+y^2z-xyz\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2+y^2-xy\right)\)
\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)
\(=0\left(x^2-xy+y^2\right)\)
\(=0\left(ĐPCM\right)\)
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\(VT=x^3+y^3+z^3-3xyz.\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)=VP\left(đpcm\right)\)
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* Có BĐT : \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ ( Chứng minh bằng xét hiệu )
Ta có BĐT : \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\Rightarrow\dfrac{x+y}{x^2+y^2}\le\dfrac{2\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{2}{x+y}\)
Chứng minh tương tự khi đó :
\(P\le\dfrac{2}{x+y}+\dfrac{2}{y+z}+\dfrac{2}{z+x}\)
\(\Rightarrow2P\le\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}=2.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=4032\)
\(\Rightarrow P\le2016\)
(x+y)3-3x2y-3xy2+z3-3xyz
=> ((x+y)3+z3)- 3xy(x+y+z)
=>(x+y+z)((x+y)2-z(x+y)+z2)-3xy(x+y+z)
=>(x+y+z)(x2+2xy+y2-xz-yz+z2-3xy)
vì x+y+z=0 => biểu thúc trên bằng 0
=> x3+y3+z3-3xyz=0
=>x3+y3+z3=3xyz
=>
=>