$a,b,c$ 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...


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

$$\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{1}{2}=\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{a+b+c}{2}=\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)=$$ $$=\sum_{cyc}\frac{a^3-ab^2}{2(a^2+b^2)}=\sum_{cyc}\left(\frac{a^3-ab^2}{2(a^2+b^2)}-\frac{a-b}{2}\right)=\sum_{cyc}\frac{(a-b)^2b}{2(a^2+b^2)}\geq0.$$ Done!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.