Question

If x,y,z>0, and x+y+z=1, then prove xy+yz+zx≥9xyz.

Collected in the board: Inequality

Steven Zheng posted 2 weeks ago

Answer 1

Use the Cauchy-Schwarz inequality

(a^2_1+a^2_2+a^2_3)(b^2_1+b^2_2+b^2_3)\geq (a_1b_1+a_2b_2+a_3b_3)^2


\dfrac{xy+yz+zx}{xyz} = \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}
(1)

Since x+y+z=1

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} =(\dfrac{1}{x}+ \dfrac{1}{y}+\dfrac{1}{z})\cdot (x+y+z)

\geq (\dfrac{1}{x}\cdot x+\dfrac{1}{y}\cdot y +\dfrac{1}{z}\cdot z ) ^2=9

Therefore

xy+yz+zx≥9xyz

Steven Zheng posted 2 weeks ago

Answer 2

Use Harmonic mean Arithmetic Mean (HM-AM) Inequality

\dfrac{3}{ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}} \leq \dfrac{x+y+z}{3}

Cross multiply

(\dfrac{1}{x}+ \dfrac{1}{y}+\dfrac{1}{z})\cdot (x+y+z)\geq 9

Since x+y+z = 1,

LHS = \dfrac{xy+yz+zx}{xyz}

Therefore

xy+yz+zx≥9xyz

Steven Zheng posted 2 weeks ago

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