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?

mình nhìn nhầm đề tưởng xyz =1 ;))))

Áp dụng AM - GM

\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2\ge2y^2\)

\(\left(\frac{xy}{z}\right)^2+\left(\frac{xz}{y}\right)^2\ge2x^2\)

\(\left(\frac{zy}{x}\right)^2+\left(\frac{zx}{y}\right)^2\ge2y^2\)

cộng vế với vế có 

\(2\left(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2\left(\frac{xz}{y}\right)^2\right)\ge\left(x^2+y^2+z.^2\right).2\ge6\)

\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\)

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