<s>[INST] 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.
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 [/INST] 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.
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