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Sa se arate ca \( \{m,a,b,c\}\subset\[0,\infty\)\ \Longrightarrow\ \prod_{\mathrm {cic}}\ \left(b^2+c^2+mbc\right)\ \ge\ abc\ \cdot\ \prod_{\mathrm {cic}}\ \left(b+c+m\sqrt {bc}\right) \) .

Remarca. Pentru m:=0 sau m:=2 se obtine inegalitatea de aici .

In general, \( \{m,n,p,a,b,c\}\subset\left[0,\infty\right)\ \Longrightarrow\ \prod_{\mathrm {cic}}\ \left(b^2+c^2+mbc\right)\ \ge\ abc\ \cdot\ \prod_{\mathrm {cic}}\ \left(b+c+\sqrt {npbc}\right) \) .

Pentru prima:

\( (b^2+c^2+mbc)(a^2+b^2+mab)\ge (ba+cb+mb\sqrt{ac})^2=b^2(a+c+m\sqrt{ac})^2 \) si analoagele, apoi le inmultim.

Pentru a doua la fel.