Câu 2: \(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+2\left(x^2+y^2+z^2\right)\)

\(=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+6\)

Áp dụng bất đẳng thức AM - GM ta có :

\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\sqrt[3]{\left(\frac{xy}{z}\right)^2\left(\frac{yz}{x}\right)^2\left(\frac{xy}{y}\right)^2}=3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^2}}=3\)\(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge\sqrt{3+6}=3\left(dpcm\right)\)

tại sao lại suy ra đc \(3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^{^2}}}=3\) vậy cậu?

Lời giải:

Ta có:

$(a+b+c)^2-(a^2+b^2+c^2)=1-1=0$

$\Leftrightarrow 2(ab+bc+ac)=0$

$\Leftrightarrow ab+bc+ac=0$

Đặt $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=t\Rightarrow x=\frac{a}{t}, y=\frac{b}{t}, z=\frac{c}{t}$

Do đó:

$xy+yz+xz=\frac{ab}{t^2}+\frac{bc}{t^2}+\frac{ac}{t^2}$

$=\frac{1}{t^2}(ab+bc+ac)=\frac{1}{t^2}.0=0$

Ta có đpcm.

cảm ơn

2: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{xy+yz+xz}{xyz}=0\)

=>xy+yz+xz=0

=>xy=-xz-yz; yz=-xy-xz; xz=-xy-yz

\(x^2+2yz=x^2+yz+yz\)

\(=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

\(y^2+2xz=y^2+xz+xz\)

\(=y^2+xz-xy-yz=y^2-xy+xz-yz\)

=y(y-x)+z(x-y)

=z(x-y)-y(x-y)=(x-y)(z-y)

\(z^2+2xy\)

\(=z^2+xy+xy\)

\(=z^2+xy-yz-xz\)

\(=z^2-xz+xy-yz=z\left(z-x\right)+y\left(x-z\right)=\left(x-z\right)\left(y-z\right)\)

\(A=\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)

\(=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{y^2z-yz^2-x^2z+xz^2+x^2y-xy^2}{\left(x-y\right)\cdot\left(x-z\right)\left(y-z\right)}\)

\(=\frac{z\left(y^2-x^2\right)+z^2\left(x-y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left\lbrack-z\left(x+y\right)+z^2+xy\right\rbrack}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(=\frac{-xz-yz+z^2+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z^2-yz-xz+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z\left(z-y\right)-x\left(z-y\right)}{\left(x-z\right)\left(y-z\right)}=\frac{\left(z-x\right)\left(z-y\right)}{\left(z-x\right)\left(z-y\right)}\)

=1