Question

[Phương Minh Trần]

Cho x,y,z >0

CM Xy/z+xz/y+zy/x>=z+y+z

Giúp minh với

Answer 1

[Phùng Khánh Linh]

Áp dụng BĐT : \(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 2 ( a > 0 ; b > 0)

Ta có : \(\dfrac{xy}{z}+\dfrac{xz}{y}\) = \(x\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\) ≥ 2x ( x > 0 ; y > 0 ; z > 0) (1)

\(\dfrac{xz}{y}+\dfrac{zy}{x}=z\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\) ≥ 2z ( x > 0 ; y > 0 ; z > 0) ( 2)

\(\dfrac{xy}{z}+\dfrac{zy}{x}=y\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\) ≥ 2y ( x > 0 ; y > 0 ; z > 0) ( 3)

Cộng từng vế của ( 1 ; 2 ; 3)

⇒\(\dfrac{xy}{z}+\dfrac{xz}{y}\) + \(\dfrac{xz}{y}+\dfrac{zy}{x}\) + \(\dfrac{xy}{z}+\dfrac{zy}{x}\) ≥ 2x + 2y + 2z
⇔ \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{zy}{x}\) ≥ x + y + z

Answer 2

[Ma Sói]

Dễ thôi

\(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\)

\(xyz(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y})\ge xyz(x+y+z)\)

\(x^2y^2+x^2z^2+y^2z^2\ge x^2yz+xz^2y+y^2zx\)\(2x^2y^2+2x^2z^2+2y^2z^2\ge2x^2yz+2xz^2y+2y^2zx\)

\((x^2y^2-2x^2yz+x^2z^2)+(y^2z^2-2y^2zx+x^2y^2)+(x^2z^2-2yz^2x+y^2z^2)\ge0\)

\(\left(xy-xz\right)^2+\left(xz-yz\right)^2+\left(yz-xy\right)^2\ge0\left(lđ\right)\)