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Prove that if $a,b,c$ are positive real numbers then:

$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \le \frac{3}{2}\frac{a+b+c}{ab+bc+ca}$

Is this true and how can we prove it? I guess it should be something relatively easy.
I came across this while working on another problem here:
Inequality using Cauchy-Schwarz

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peter.petrov
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1 Answers1

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it is equivalent to $$(a^2-c^2)(a-c)b+(a-b)(a^2-b^2)c+a(b^2-c^2)(b-c)\geq 0$$ which is positive.

Dr. Sonnhard Graubner
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  • Thanks, I will verify your calculations. Thanks a lot. – peter.petrov Mar 15 '16 at 15:48
  • OK, it is this indeed. I got: $a^3 b + a^3 c + b^3 a + b^3 c + c^3 a + c^3 b - 2 a^2 b c - 2 a b^2 c - 2 a b c^2 $ which is the same as your expression. But the representation you showed is the key. No idea how you found this so quickly, it took me about 2 hours just to get $a^3 b + a^3 c + b^3 a + b^3 c + c^3 a + c^3 b - 2 a^2 b c - 2 a b^2 c - 2 a b c^2 $. And even from here I am not sure I would have been able to come up with your representation. – peter.petrov Mar 15 '16 at 18:08