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