What tools one should use for inequalities? If $a,b,c>0$ prove that:
$$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$
My first try was the following:
$$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\sqrt[3]{16abc}}=\frac{1}{\sqrt[3]{16abc}}$$
But $\frac{1}{\sqrt[3]{16abc}}\geq \frac{1}{3\sqrt[3]{abc}}$
The I have tried your method from another post:
$$\sum_{cyc}\frac{1}{a+4b+4c}=\sum_{cyc}\frac{1}{a+2b+2(b+2c)}$$
$$\sum_{cyc}\frac{1}{a+2b+2(b+2c)}\leq\sum_{cyc}\frac{1}{9}\left (\frac{1^2}{a+2b}+\frac{2^2}{b+2c}  \right )$$
Where I got $$\sum_{cyc}\frac{1}{3}\left ( \frac{1}{a+2b} \right )\leq \frac{1}{3\sqrt[3]{abc}}$$ wich is false.
$$$$
 A: This is sort of an ugly proof which uses Maclaurin inequalities.
Define constants $A,B,C, \alpha,\beta,\gamma$ through following polynomial:
$$P(\lambda) = (\lambda-a)(\lambda-b)(\lambda-c)
  = \lambda^3 - A\lambda^2 + B\lambda - C
  = \lambda^3 - 3\alpha\lambda^2 + 3\beta^2\lambda - \gamma^3$$
By Vieta's formulas, we have 
$$a + b + c = A = 3\alpha\quad\text{ and }\quad abc = C = \gamma^3$$
The LHS of the inequality at hand can be rewritten as
$$\begin{align}{\rm LHS} &= \sum_{cyc} \frac{1}{4A - 3a}
= \frac13 \sum_{cyc}\frac{1}{4\alpha-a}
= \frac13 \frac{P'(4\alpha)}{P(4\alpha)}
= \left.\frac{\lambda^2-2\alpha\lambda+\beta^2}{\lambda^3-3\alpha\lambda^2+3
\beta^2\lambda - \gamma^3}\right|_{\lambda=4\alpha}\\
&= \frac{8\alpha^2+\beta^2}{16\alpha^3 + 12\alpha\beta^2 - \gamma^3}
\end{align}
$$
while the RHS equals to $\displaystyle\;\frac{1}{3\gamma}$. The inequality we want to prove
is equivalent to
$$\begin{align}
{\rm LHS} \stackrel{?}{\le} {\rm RHS}
\iff & \frac{8\alpha^2+\beta^2}{16\alpha^3 + 12\alpha\beta^2 - \gamma^3} \stackrel{?}{ \le} \frac{1}{3\gamma}\\
\iff &
16\alpha^3 + 12\alpha\beta^2 - \gamma^3 - 3\gamma(8\alpha^2 + \beta^2) \stackrel{?}{\ge} 0
\end{align}\tag{*1}
$$
By Maclaurin's inequality, we have $\alpha \ge \beta \ge \gamma$.
This implies
$$\begin{align} 
&\; 16\alpha^3 + 12\alpha\beta^2 - \gamma^3 - 3\gamma(8\alpha^2 + \beta^2)\\ 
\ge &\; 16\alpha^3 + 12\alpha\beta^2 - \beta^3 - 3\beta(8\alpha^2 + \beta^2)\\ 
 =  &\; 16\alpha^3 - 24\alpha^2\beta + 12\alpha\beta^2 - 4\beta^3\\
= &\; 4(\alpha^3 - \beta^3) + 12\alpha(\alpha-\beta)^2\\ 
\ge &\; 0
\end{align}$$
This establish the last line in $(*1)$. As a result, the inequality at hand is valid.
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(v^2)\geq0,$ where
$$f(v^2)=16u^3+12uv^2-w^3-24u^2w-3u^2w,$$ which is a linear function.
But a linear function gets a minimal value for an extreme value of $v^2$, which happens for equality case of two variables.
Sines our inequality is homogeneous, we can assume that $abc=1$ and for $b=a$ and $c=1/a^2$ we obtain:
$$\frac{2}{4/a^2+5a}+\frac{1}{\frac{1}{a^2}+8a}\leq\frac{1}{3}$$ or
$$(a-1)^2(40a^4+17a^3-6a^2+8a+4)\geq0,$$ which is obvious.
