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$a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...

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We have $\frac{ab^2}{a^2 +b^2}\leq \frac{b}{2} ,\frac{bc^2}{c^2 +b^2}\leq \frac{c}{2} ,\frac{ca^2}{c^2 +a^2}\leq \frac{a}{2}$ hence $$\frac{a^3}{a^2 +b^2} +\frac{b^3}{c^2 +b^2} +\frac{c^3}{a^2 +c^2} =a+b+c - \left(\frac{ab^2}{a^2 +b^2}+\frac{bc^2}{c^2 +b^2}+\frac{ca^2}{c^2 +a^2}\right) \geq \frac{a+b+c}{2}.$$

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  • $\begingroup$ It doesn't look concave. $\endgroup$ – Quang Hoang Sep 15 '14 at 10:51
  • $\begingroup$ Its concave only for part of the interval $(0, 1)$ and convex in the rest. Also concave means by Jensen the inequality is the other way round... $\endgroup$ – Macavity Sep 15 '14 at 10:53
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    $\begingroup$ Yes, you're right. I'd make mistake. Now, I edit and it should be ok. $\endgroup$ – user110661 Sep 15 '14 at 11:12
  • $\begingroup$ Yep - looks good now. $\endgroup$ – Macavity Sep 15 '14 at 11:14
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$$\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{1}{2}=\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{a+b+c}{2}=\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)=$$ $$=\sum_{cyc}\frac{a^3-ab^2}{2(a^2+b^2)}=\sum_{cyc}\left(\frac{a^3-ab^2}{2(a^2+b^2)}-\frac{a-b}{2}\right)=\sum_{cyc}\frac{(a-b)^2b}{2(a^2+b^2)}\geq0.$$ Done!

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