all the other questions are here: http://guas.info/competit/tuymaada/2016e.tex Results: http://guas.info/competit/tuymaada/res16.xls

rightways wrote: Non-negative numbers $a$, $b$, $c$ satisfy $a^2+b^2+c^2\geq 3$. Prove the inequality $$ (a+b+c)^3\geq 9(ab+bc+ca). $$ It's \[ 2p^3-9p^2+27 =(p-3)^2(p+3/2) \ge 0,p=a+b+c \]

we have to prove: (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca)^3 ≥ 81(ab + bc + ca)^2 by A-M G-M: (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca)^3 ≥ (3((a^2 + b^2 + c^2).(ab + bc + ca)^2)^(1/3))^3 = 81(ab + bc + ca)^2