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Algebra Difficulty 5.9 AIME, harder Prove it

Example 1 Given that x,y,zx, y, z are positive real numbers, and xy+yz+zx=1x y + y z + z x = 1, prove:
(x+y)(y+z)(z+x)≥89(x+y+z)(x+y)(y+z)(z+x) \geq \frac{8}{9}(x+y+z)

Solution

Proof: By the AM-GM inequality for three variables, we have
x+y+z=(x+y+z)⋅(xy+yz+zx)≥3xyz3⋅3(xyz)23=9xyzx+y+z=(x+y+z) \cdot(x y+y z+z x) \geq 3 \sqrt[3]{x y z} \cdot 3 \sqrt[3]{(x y z)^{2}}=9 x y z
 i.e., 9xyz≤19(x+y+z) , \text { i.e., } 9 x y z \leq \frac{1}{9}(x+y+z) \text { , }

Thus, by the identity (*), we get
(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)−xyz=(x+y+z)−xyz≥(x+y+z)−19(x+y+z)=89(x+y+z)\begin{aligned} (x+y)(y+z)(z+x) & =(x+y+z)(x y+y z+z x)-x y z \\ & =(x+y+z)-x y z \\ & \geq(x+y+z)-\frac{1}{9}(x+y+z) \\ & =\frac{8}{9}(x+y+z) \end{aligned}

Therefore, (x+y)(y+z)(z+x)≥89(x+y+z)(x+y)(y+z)(z+x) \geq \frac{8}{9}(x+y+z).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.