For $a,b,c>0$ prove the inequality $$ (1+a+ab)(1+b+bc)(1+c+ca) \leq (1+a+a^2)(1+b+b^2)(1+c+c^2). $$ I know that I should use the multiplicative rearrangement inequality but I am not sure how to choose the involved sequences correctly. Any ideas?
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For the rearrangement inequality see the question here. – Dietrich Burde Jun 10 '15 at 18:08
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Thank you but that is my question :) – Leox Jun 10 '15 at 18:16
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HINT: use that $$x^2+y^2+z^2\geq xy+xz+yz$$ for all real $x,y,z$ is hold. And your inequality can be written as $$(ab)^2+(ac)^2+(bc)^2-a^2bc-ab^2c-abc^2+a^2b+ac^2+b^2c-3abc+a^2+b^2+c^2-ab-ac-bc\geq 0$$
Dr. Sonnhard Graubner
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Your decomposition is rather unclear. I see that $(ab)^2+(ac)^2+(bc)^2-a^2bc-ab^2c-abc^2 \geq 0$ and $a^2+b^2+c^2-ab-ac-bc\geq 0$, but what do you do with $a^2b+ac^2+b^2c-3abc$ ? – Ewan Delanoy Jun 10 '15 at 19:14
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this is AM-GM, since we have $$\frac{a^2b+ac^2+b^2c}{3}\geq \sqrt[3]{a^3b^3c^3}=abc$$ – Dr. Sonnhard Graubner Jun 10 '15 at 19:22