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mso-style-noshow:yes;
mso-style-priority:99;
mso-style-parent:"";
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mso-para-margin-top:0in;
mso-para-margin-right:0in;
mso-para-margin-bottom:10.0pt;
mso-para-margin-left:0in;
line-height:115%;
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mso-ascii-theme-font:minor-latin;
mso-hansi-font-family:Calibri;
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x3+y3=(x+y)3-3xy(x+y)$\geqslant(x+y)^{3}-\frac{3}{4}(x+y)^{3}$=$\frac{(x+y)^{3}}{4}$
4(x3+y3)$\geqslant $(x+y)3 tuong tu
4(y3+z3)$\geqslant $(y+x)3 ;4(x3+z3)$\geqslant $(x+z)3
$\sqrt[3]{4(x^{3}+y^{3})}$+$\sqrt[3]{4(x^{3}+z^{3})}$+$\sqrt[3]{4(y^{3}+z^{3})}$$\geqslant $6$\sqrt[3]{xyz}$$2(\frac{x}{y^{2}}+\frac{y}{z^{2}}+\frac{z}{x^{2}})\geqslant \frac{6}{\sqrt[3]{xyz}}$$P\geqslant 6(\sqrt[3]{xyz}+\frac{1}{\sqrt[3]{xyz}})\geqslant 12$
Pmin =12 khi x=y=z=1
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