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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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    $\begingroup$ You can use the inequality of arithmetic and geometric means (short: AMGM-inequality). It says that $\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1 x_2 ... x_n}$. $\endgroup$ – Tim Dikland Jul 1 '18 at 12:53
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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.

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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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