Notice that $3(a^3+b^3+c^3)\ge(a^2+b^2+c^2)(a+b+c)$ or $3\ge(a+b+c)$.

darn,why such a easy problem on a TST? solution = by tchebycheff we have $3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2) \Longrightarrow 3\ge a+b+c$ we are done!

we have that: \[x^3-x^2-x+1=(x-1)^2(x+1) \geq 0\] Then: \[ a+b+c - 3 = \sum (a-1) \leq \sum (a^3-a^2)=0\]

I have two solutions for this?First by Chebishev.Second by C-S we have $(a^3+b^3+c^3)(a+b+c)\geq (a^2+b^2+c^2)^2$ or $a+b+c\geq a^2+b^2+c^2\geq 2a+2b+2c-3$ or $a+b+c\leq 3$

socrates wrote: Find the maximum possible value of $a + b + c ,$ if $a,b,c$ are positive real numbers such that $a^2 + b^2 + c^2 = a^3 + b^3 + c^3 .$ Bulgaria JTST 2014