Rijul saini wrote: Indian Postal Training Set 2, Problem 5 For any positive real numbers $a, b, c$, prove that \[\sum_{cyclic} \frac{(b + c)(a^4 - b^2 c^2 )}{ab + 2bc + ca} \ge 0\] Note that the sequences $\left\{a^3(ab+bc+ca-bc); b^3(ab+bc+ca-ca); c^3(ab+bc+ca-ab)\right\}$ and $\left\{\frac 1{ab+bc+ca+bc}, \frac 1{ab+bc+ca+ca},\frac 1{ab+bc+ca+ab}\right\}$ are similarly sorted. Therefore applying the Rearrangement inequality it is a one-liner, \[\sum_{cyc}\frac{a^4(b+c)}{ab+2bc+ca}\geq \sum_{cyc}\frac{b^2c^2(b+c)}{ab+2bc+ca}.\] Hence proved. $\Box$

See here.

It's fifth problem of Turkey NMO 2009, also see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2003335&sid=6ac5f3f753fc1fa1b2df1a8789382121#p2003335

Potla wrote: Note that the sequences $\left\{a^3(ab+bc+ca-bc); b^3(ab+bc+ca-ca); c^3(ab+bc+ca-ab)\right\}$ and $\left\{\frac 1{ab+bc+ca+bc}, \frac 1{ab+bc+ca+ca},\frac 1{ab+bc+ca+ab}\right\}$ are similarly sorted. Therefore applying the Rearrangement inequality it is a one-liner, \[\sum_{cyc}\frac{a^4(b+c)}{ab+2bc+ca}\geq \sum_{cyc}\frac{b^2c^2(b+c)}{ab+2bc+ca}.\] Hence proved. $\Box$ Please Potla could you fill in the detail after stating it is a `one-liner' to show how you get to the end result. Thanks Merlin

Rijul saini wrote: For any positive real numbers $a, b, c$, prove that \[\sum_{cyclic} \frac{(b + c)(a^4 - b^2 c^2 )}{ab + 2bc + ca} \ge 0\] 1st proof $ab+2bc+ca \leq (b+c)(a+\sqrt{bc}) (1)$ this is true Proof. After expanding we will have to prove $2bc \leq b\sqrt{bc}+c\sqrt{bc}$ which is easy by A.M G.M Now after we use $(1)$ on our ineq it will become $\sum a^3+3abc \geq \sum a^2\sqrt{bc}+ \sum ab\sqrt{ab}$ which can be proved by Muirhead