Cho x+y+z=a $\frac{1}{x}$+$\frac{1}{y}$+ $\frac{1}{z}$= $\frac{1}{a}$ CMR: trong 3 số x,y,z ít nhất 1 số bằng a

ĐK: $x\ne 0,y\ne 0,z\ne 0, x+y+z\ne 0$

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{a}\\x+y+z=a\to \dfrac{1}{x+y+z}=\dfrac{1}{a}\\\to \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\\\to \dfrac{xy+yz+xz}{xyz}=\dfrac{1}{x+y+z}\\\to (xy+yz+xz)(x+y+z)-xyz=0\\\to x^2y +xy^2+xyz + xyz + y^2z +yz^2 +x^2z+xyz +xz^2-xyz=0\\\to x^2y +xy^2 + y^2z+yz^2 +x^2z+xz^2+2xyz=0\\\to xy(x+y) + z^2(x+y)+yz(x+y) + xz(x+y)=0\\\to (x+y)(xy+z^2+yz+xz)=0\\\to (x+y)(x(y+z)+z(y+z))=0\\\to (x+y)(y+z)(x+z)=0\\\to \left[ \begin{array}{l}x+y=0\\x+z=0\\y+z=0\end{array} \right.$

Không mất tính tổng quát giả sử $x+y=0$

$\to z=a$ (Đpcm)