Inequality of three variable polynomial I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that
$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
How this can be done? Is it possible to prove stronger inequalities like
$$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$
or even 
$$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
without computer?
 A: You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS.  A naive way to do this such as
$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$
uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality
$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$
after cyclic summation.  A slightly better combination of terms is
$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$
but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality
$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$
after cyclic summation.  You get a strong enough result by mixing some of the $40$ terms with each other to get
$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$
which proves
$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$
Now multiply by $10$ and apply AM-GM to the remaining terms to conclude.  The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.
A: I don't have enough karma to comment, so I'll just drop this here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225
discusses the problem.
A: Each of the inequalities below is easily verified by AM-GM
\begin{eqnarray*}
&29(& a^6 &     &     & + a^3bc^2 & +3a^2b^3c &           &          &          &          & \geq & 5a^3b^2c &          &          &) \\
&29(&     & b^6 &     &           & + a^2b^3c & +3ab^2c^3 &          &          &          & \geq &          & 5ab^3c^2 &          &) \\
&29(&     &     & c^6 & +3a^2b^3c &           & + ab^2c^3 &          &          &          & \geq &          &          & 5a^2bc^3 &) \\
&4 (& a^6 &     &     &           & + a^2b^3c &           &          & +ab^3c^2 &          & \geq & 3a^3b^2c &          &          &) \\
&4 (&     & b^6 &     &           &           & + ab^2c^3 &          &          & +a^2bc^3 & \geq &          & 3ab^3c^2 &          &) \\
&4 (&     &     & c^6 & + a^3bc^2 &           &           & +a^3b^2c &          &          & \geq &          &          & 3a^2bc^3 &). \\
\end{eqnarray*}
Now add these together, cancel some terms, divide by $3$ and we are dumb.
