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If $a+b+c=1$ and a,b,c>0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?


marked as duplicate by Martin R, Macavity inequality Mar 22 at 8:52

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  • 1
    $\begingroup$ You asked the identical question three years ago (and got an answer) ... $\endgroup$ – Martin R Mar 22 at 8:39

It's wrong for real numbers.

For positive variables by C-S $$\sum_{cyc}\frac{a^2}{a^2+c}=\sum_{cyc}\frac{a^2(a+b)^2}{(c(a+b+c)+a^2)(a+b)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2}.$$ Thus, it's enough to prove that $$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2$$ or $$\sum_{cyc}(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)\geq0,$$ which is true because $$\sum_{cyc}(a^4-a^3b)\geq0$$ by Rearrangement; $$\sum_{cyc}a^3c\geq\sum_{cyc}a^2bc$$ it's $$\sum_{cyc}\frac{a^2}{b}\geq\sum_{cyc}a,$$ which is true by Rearrangement again and $$\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$

  • $\begingroup$ +1 for the good efforts, even though duplicate. $\endgroup$ – Macavity Mar 22 at 8:53
  • $\begingroup$ @Macavity: Michael already answered the original question: math.stackexchange.com/a/1602340/42969 :) $\endgroup$ – Martin R Mar 22 at 8:54
  • $\begingroup$ @Macavity Thank you! I usually don't remember what I proved. $\endgroup$ – Michael Rozenberg Mar 22 at 9:36
  • $\begingroup$ @MichaelRozenberg: It is quite easy with Approach0 – it takes less than a minute to find that this question has been asked and answered before. $\endgroup$ – Martin R Mar 22 at 10:06

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