<s>[INST] Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers I was just encountered an inequality in AoPs, Here it is: 
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=4569&view=next, 
that is: 

If $a$, $b$ and $c$ are positive numbers, then we have inequality: $$\dfrac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{8}\ge\dfrac{(ab+bc+ca)^3}{27}$$

If the above ineq is succeed, it's easy to generalize the variables into real number. 
Here I've got an terrible solution:
just expand(I think it is complicated to deal with this ineq using expand and easy to make mistake) and the inequality is equivalent to the following symmetric Ineq.
$$6a^2b^2c^2+27\sum_{sym} a^4b^2\ge4\sum_{sym} a^3b^3+24\sum_{sym} a^3b^2c,$$
then by the A-G Ineq, 
$$6a^2b^2c^2+\sum\limits_{sym} b^4c^2=\sum\limits_{sym}(a^4b^2+a^2b^2c^2)\ge2\sum\limits_{sym}a^3b^2c, $$
and Muirhead's theorem implies
\begin{equation}
\sum_{sym}a^4b^2\ge\sum_{sym}a^3b^2c,\quad
\sum_{sym}a^4b^2\ge\sum_{sym}a^3b^3.
\end{equation}
finally the ineq can be proved by three appropriate coeffients multiplied to the previous three ineq and total them up. 
Is there a concise proof without complicated calculating?(also it swept the buffalo way away). this is what I interested. 
 [/INST] We'll prove that your inequality is true for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+b^2)=\prod\limits_{cyc}(a^2+b^2+c^2-c^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$, 
we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function,
which says that it's enough to prove our inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Since our inequality is even degree, it's enough to assume $b=c=1$,
which gives $27(a^2+1)^2\geq4(2a+1)^3$ or $(a-1)^2(27a^2+22a+23)\geq0$.
Done! 
</s>