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Is it possible to prove that $ \dfrac {a}{a+b}+\dfrac {b}{b+c}+\dfrac {c}{c+a}\geq \dfrac {3}{2}$, if $ a+b+c=1, a,b,c>0$?

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  • $\begingroup$ Actually the problem was that: if a,b,c are real, positive numbers,a+b+c=1 prove that $\dfrac {a^{2}+b}{b+c}+\dfrac {b^{2}+c}{c+a}+\dfrac {c^{2}+a}{a+b}\geq 2$ $\endgroup$ – Brain123 Feb 6 '18 at 19:12
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I think it's wrong.

Try $c\rightarrow0^+$.

We obtain $$\frac{a}{a+b}\geq\frac{1}{2}$$ or $$a\geq b$$ and you can get a counterexample with $a<b$.

Also, we can see it for $a\rightarrow0^+$: we obtain $0\geq\frac{1}{2}.$

By the way, your second problem is true.

Indeed, we need to prove that $$\sum_{cyc}\frac{a^2+b}{b+c}\geq2$$ or $$\sum_{cyc}(a^2+b)(a+b)(a+c)\geq2\prod_{cyc}(a+b)$$ or $$\sum_{cyc}(a^2+b)(a^2+ab+ac+bc)\geq2\prod_{cyc}(a+b)$$ or $$\sum_{cyc}(a^2+b)(a+bc)\geq2\prod_{cyc}(a+b)$$ or $$\sum_{cyc}\left(a^3+\frac{1}{3}abc+ab+a^2b\right)\geq2\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)$$ or $$\sum_{cyc}\left(a^3+\frac{1}{3}abc+a^2b+a^2c+abc+a^2b\right)\geq2\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)$$ or $$\sum_{cyc}(a^3-a^2c)\geq0,$$ which is true by Rearrangement.

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  • $\begingroup$ +1 I suspect @gimusi was correct and the OP missstated Nesbitt's inequality. $\endgroup$ – almagest Feb 6 '18 at 19:05
  • $\begingroup$ Yes, it's not Nesbitt. $\endgroup$ – Michael Rozenberg Feb 6 '18 at 19:07
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Michael is right, computing the left hand side minus the right hand side we obtain $$1/2\,{\frac { \left( b-c \right) \left( a-c \right) \left( a-b \right) }{ \left( a+b \right) \left( b+c \right) \left( c+a \right) }} \geq 0$$

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