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
\(2\left(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2\left(\frac{xz}{y}\right)^2\right)\ge\left(x^2+y^2+z.^2\right).2\ge6\)
\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\)