$x^2+y^2+z^2+ mxy + nxz + pyz \geq 6(\sqrt[3]{2m^2}+\sqrt[3]{2n^2}+\sqrt[3]{2p^2})$ because $z^2+mxy \geq 3\sqrt[3]{z^2\frac{(mxy)^2}{4}}=3\sqrt[3]{\frac{8^2 m^2}{4}}=6\sqrt[3]{2m^2}$ and equality for $z=\sqrt[3]{4m}$

I will show only part b. since a. follows immediately from b. \begin{align*} x^2+y^2+z^2+mxy+nxz+pyz & \geq \left(3\sum_{cyc}{ p^{2/3}}\right)\left( \sqrt[3(p^{2/3}+q^{2/3}+r^{2/3})]{\prod_{cyc}{\left( \frac{x^2}{p^{2/3}}\right) ^{p^{2/3}}}\cdot \prod_{cyc}{\left( \frac{mxy}{2\cdot m^{2/3}}\right) ^{2\cdot m^{2/3}}}}\right)\\ & =\left(3\sum_{cyc}{ p^{2/3}}\right) \sqrt[3]{\prod_{cyc}{\frac{x^2}{4}}} \\ & = 6\sqrt[3]{2}\sum_{cyc}{m^{2/3}} \\ \end{align*}Equality holds at $x=\sqrt[3]{4p},y=\sqrt[3]{4n},z=\sqrt[3]{4m}$.

ThE-dArK-lOrD wrote: I will show only part b. since a. is follow immediately from b. By Weight A.M.-G.M., we have $x^2+y^2+z^2+mxy+nxz+pyz\geq \Big(\sum_{cyc}{\Big( \sqrt[3]{p^2}+2\sqrt[3]{m^2}\Big)}\Big) \sqrt[(\sum_{cyc}{( \sqrt[3]{p^2}+2\sqrt[3]{m^2})})]{ \prod_{cyc}{\Big( \frac{x^2}{\sqrt[3]{p^2}}\Big) ^{\sqrt[3]{p^2}}}\times \prod_{cyc}{\Big( \frac{mxy}{2\sqrt[3]{m^2}}\Big) ^{2\sqrt[3]{m^2}}}}=3\sum_{cyc}{\sqrt[3]{p^2}}\sqrt[(3\sum_{cyc}{\sqrt[3]{p^2}})]{\prod_{cyc}{\Big( x^{2\sqrt[3]{p^2}+2\sqrt[3]{m^2}+2\sqrt[3]{n^2}}\Big)}\prod_{cyc}{\Big( \frac{(\frac{\sqrt[3]{m}}{2})^{2\sqrt[3]{m^2}}}{(\sqrt[3]{p^2})^{\sqrt[3]{p^2}}} \Big)}}=3\sum_{cyc}{\sqrt[3]{p^2}}\sqrt[3\sum_{cyc}{\sqrt[3]{p^2}}]{\Big( \prod_{cyc}{\frac{x^2}{4}}\Big)^{\sum_{cyc}{\sqrt[3]{m^2}}}}=3\Big( \sum_{cyc}{\sqrt[3]{p^2}}\Big) \sqrt[3]{\prod_{cyc}{\frac{x^2}{4}}}=6\sqrt[3]{2}\sum_{cyc}{\sqrt[3]{m^2}}$ Equality hold at $x=\sqrt[3]{4p},y=\sqrt[3]{4n},z=\sqrt[3]{4m}$ Radical's order must be a natural number and, in your solution, it clearly isn't.

Pure_IQ wrote: Radical's order must be a natural number and, in your solution, it clearly isn't. Why so? Weighted AM-GM is valid for all positive real weights.

Ankoganit wrote: Pure_IQ wrote: Radical's order must be a natural number and, in your solution, it clearly isn't. Why so? Weighted AM-GM is valid for all positive real weights. I thought that the weights must have the sum 1 ,in order that radical's order be 1 too. I've learned that radical's order must be a natural number. Thank you !

Weighted Form[edit] The weighted form of AM-GM is given by using weighted averages. For example, the weighted arithmetic mean of $x$ and $y$ with $3:1$ is $\frac{3x+1y}{3+1}$ and the geometric is $\sqrt[3+1]{x^3y}$. AM-GM applies to weighted averages. Specifically, the weighted AM-GM Inequality states that if $a_1, a_2, \dotsc, a_n$ are nonnegative real numbers, and $\lambda_1, \lambda_2, \dotsc, \lambda_n$ are nonnegative real numbers (the "weights") which sum to 1, then \[\lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n},\]or, in more compact notation, \[\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} .\]Equality holds if and only if $a_i = a_j$ for all integers $i, j$ such that $\lambda_i \neq 0$ and $\lambda_j \neq 0$. We obtain the unweighted form of AM-GM by setting $\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n$. I am confuse.

@PureIQ The basic form of weighted AM-GM involves weights summing to $1$. But we can easily generalise it to any positive sum of weights.

I know how it can be generalised, i just was wondering about that radical's order ,because till now, i thought it can't be rational or irational. Now is ok .