See here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=508673&p=2863558

$\sum \frac{a}{a^2+b^3+c^3}=\sum \frac{a^2}{a^3+ab^3+ac^3}$ $\geq \frac{(a+b+c)^2}{a^3+b^3+c^3+\sum_{sym}a^3b}$ $= \frac{(a+b+c)^2}{a^4+b^4+c^4+\sum_{sym}a^3b}$ $= \frac{(a+b+c)^2}{(a+b+c)(a^3+b^3+c^3)}$ $= \frac{a+b+c}{a^3+b^3+c^3}$ $= \frac{(a+b+c)(a^4+b^4+c^4)^2}{(a^3+b^3+c^3)^3}$ But the last expression is obviously $\geq 1$ by Holder. Thus, we are done.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=520657 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=571748&p=3359441#p3359441

Never mind

Is there any solution beside holder inequality at the last point?

$/bump$ .....

Since $a,b,c$ are the positive, then we can use Titu's lemma: $\sum_{cyc} \frac{a}{a^2+b^3+c^3}=\sum_{cyc} \frac{a^2}{a^3+ab^3+ac^3}$ $\geq \frac{(a+b+c)^2}{a^3+b^3+c^3+\sum_{sym}a^3b}=\frac{(a+b+c)^2}{(a+b+c)(a^3+b^3+c^3)}=\frac{(a+b+c)}{(a^3+b^3+c^3)}$ Now it is enough to show that $\frac{(a+b+c)}{a^3+b^3+c^3} \geq 1$, which can be rewritten as $(a+b+c)\geq a^3+b^3+c^3$ We should there homogenize both sides using $a^3+b^3+c^3=a^4+b^4+c^4$ $(a+b+c)(a^4+b^4+c^4)(a^4+b^4+c^4) \geq (a^3+b^3+c^3)^3$ which is true by Holder's inequality

Bakhtier wrote: Since $a,b,c$ are the positive, then we can use Titu's lemma: $\sum_{cyc} \frac{a}{a^2+b^3+c^3}=\sum_{cyc} \frac{a^2}{a^3+ab^3+ac^3}$ $\geq \frac{(a+b+c)^2}{a^3+b^3+c^3+\sum_{sym}a^3b}=\frac{(a+b+c)^2}{(a+b+c)(a^3+b^3+c^3)}=\frac{(a+b+c)}{(a^3+b^3+c^3)}$ Now it is enough to show that $\frac{(a+b+c)}{a^3+b^3+c^3} \geq 1$, which can be rewritten as $(a+b+c)\geq a^3+b^3+c^3$ We should there homogenize both sides using $a^3+b^3+c^3=a^4+b^4+c^4$ $(a+b+c)(a^4+b^4+c^4)(a^4+b^4+c^4) \geq (a^3+b^3+c^3)^3$ which is true by Holder's inequality We can solve the last part by Cauchy Schwarz inequality as well. a+b+c>=a^3+b^3+c^3 , which is also (a+b+c)(a^3+b^3+c^3)(a^4+b^4+c^4)^2 >= (a^3+b^3+c^3)^4 By Cauchy Schwarz : (a+b+c)(a^3+b^3+c^3)>=(a^2+b^2+c^2)^2 and (a^2+b^2+c^2)(a^4+b^4+c^4)>=(a^3+b^3+c^3)^2.

Multiply both sides by $\sum_\mathrm{cyc} (a^3+ab^3+ac^3)=(a+b+c)(a^3+b^3+c^3)$, where the equality comes from the condition. By Cauchy-Schwarz, it then suffices to show that $$(a+b+c)^2 \geq (a+b+c)(a^3+b^3+c^3) \iff (a+b+c)(a^4+b^4+c^4)^2 \geq (a^3+b^3+c^3)^3,$$which is just Holder. $\blacksquare$

Write $$\sum \frac{a^2}{a^3+ab^3+ac^3} \geq \frac{(a+b+c)^2}{a^4+b^4+c^4+\sum ab^3} = \frac{a+b+c}{a^3+b^3+c^3} \geq \left(\frac{a^3+b^3+c^3}{a^4+b^4+c^4}\right)^2$$by Holder.

crazyfehmy wrote: Let $a, b, c$ be positive real numbers satisfying $a^3+b^3+c^3=a^4+b^4+c^4$. Show that \[ \frac{a}{a^2+b^3+c^3}+\frac{b}{a^3+b^2+c^3}+\frac{c}{a^3+b^3+c^2} \geq 1 \] https://artofproblemsolving.com/community/c6h508673p2858621 https://artofproblemsolving.com/community/c6h1161640p5532866 https://artofproblemsolving.com/community/c6h1941924p13374908

The LHS can be written as \[\sum_{\text{cyc}} \frac{a^2}{a^3+ab^3+ac^3} \ge \frac{(a+b+c)^2}{a^3+b^3+c^3+ab^3+ac^3+ba^3+bc^3+ca^3+cb^3}\] using Cauchy. After substituting $a^4+b^4+c^4 = a^3+b^3+c^3$, we can cancel $a+b+c$ from the numerator and denominator to get \[\frac{a+b+c}{a^3+b^3+c^3} = \frac{(a+b+c)(a^4+b^4+c^4)^2}{(a^3+b^3+c^3)^3}.\] To finish, we note by Holder's that \[\left(\sum_{\text{cyc}} a\right)^{1/3} \left(\sum_{\text{cyc}} a^4\right)^{2/3} \ge \left(\sum_{\text{cyc}} a^{1 \cdot 1/3 + 4 \cdot 2/3}\right)^3.\]