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For $ a, b $ and $ c $ positive reals, prove that: $$\frac{a+ \sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a\left(\frac{a+b}{2}\right)\left(\frac{a+b+c}{3}\right)}$$

Solution: By AM-GM and Hölder $RHS=\sqrt[3]{\frac{a+a+a}{3}\cdot\frac{a+\frac{a+b}{2}+b}{3}\cdot\frac{a+b+c}{3}}\ge\frac{1}{3}\sqrt[3]{(a+a+a)(a+\sqrt{ab}+b)(a+b+c)}\ge LHS$

How to solve only by inequality of means?

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After replacing $a$ on $a^6$, $b$ on $b^6$ and $c$ on $c^6$ we need to prove that $$\sqrt[3]{\frac{a^6(a^6+b^6)(a^6+b^6+c^6)}{6}}\geq\frac{a^6+a^3b^3+a^2b^2c^2}{3}$$ or $$\sqrt[3]{\frac{(a^6+b^6)(a^6+b^6+c^6)}{6}}\geq\frac{a^4+ab^3+b^2c^2}{3},$$ which is true by AM-GM, Power Mean inequality and C-S: $$\sqrt[3]{\frac{(a^6+b^6)(a^6+b^6+c^6)}{6}}\geq\sqrt[3]{\frac{a^6+a^3b^3+b^6}{3}\cdot\frac{a^6+b^6+c^6}{3}}\geq$$ $$\geq\sqrt{\frac{a^4+a^2b^2+b^4}{3}\cdot\frac{a^4+b^4+c^4}{3}}\geq\sqrt{\frac{(a^4+ab^3+b^2c^2)^2}{9}}=\frac{a^4+ab^3+b^2c^2}{3}.$$

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The following quantity is clearly positive \begin{eqnarray*} 3a^6(a^3-b^3)^2+ 4a^4(a^4-b^2c^2)^2+ 3b^6(a^3-b^3)^2+ 6b^4(b^4-a^2c^2)^2+ 3a^2c^2(b^4-a^2c^2)^2+ 2^6c^2(ab-c^2)^2+ 4c^6(a^3-b^3)^2+ 3b^6c^2(ab-c^2)^2+ 4a^3b(b^4-a^2c^2)^2+ 6a^4b^4(c^2-ab)^2+ 8a^3b^3c^2(c^2-ab)^2. \end{eqnarray*} This can be rearranged to \begin{eqnarray*} 7a^{12}-6a^9b^3+12a^6b^6-2a^3b^9+9b^{12} -6a^8b^2c^2-12a^5b^5c^2-6a^2b^8c^2-6a^4b^4c^4-6ab^6c^5+9a^6c^6+7b^6c^6. \end{eqnarray*} Now setting this to be greater than or equal to zero & rearranging some more and we have

\begin{eqnarray*} 2(a^4+ab^3+b^2c^2)^3 \leq 9(a^6+b^6)(a^6+b^6+c^6). \end{eqnarray*} Multiply by $a^6$, divide by $54$ and take the cube root \begin{eqnarray*} \frac{ a^6+a^3b^3+a^2b^2c^2 }{3} \leq \sqrt[3]{ a^6 \left( \frac{a^6+b^6}{2} \right) \left(\frac{a^6+b^6+c^6}{3} \right)}. \end{eqnarray*} The result then follows by replacing $a^6 \rightarrow a$ , $b^6 \rightarrow b$ & $c^6 \rightarrow c$.

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