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Sa se arate ca

\( 9(a^3+3b^3+5c^3)\ge (a^2+3b^2+5c^2)(a+3b+5c) \)

oricare ar fi numerele reale pozitive a, b, c.

M. Chirciu, O. Stroe, G.M.

Cu inegalitatea Cauchy-Buniakowski-Schwarz avem:
\( 9\left(a^{3}+3b^{3}+5c^{3}\right)\left(a+3b+5c\right)\ge9\left(a^{2}+3b^{2}+5c^{2}\right)^{2}\Leftrightarrow9\left(a^{3}+3b^{3}+5c^{2}\right)\ge\frac{9\left(a^{2}+3b^{2}+5c^{2}\right)^{2}}{a+3b+5c} \).
Mai ramane de aratat ca
\( \frac{9\left(a^{2}+3b^{2}+5c^{2}\right)^{2}}{a+3b+5c}\ge\left(a^{2}+3b^{2}+5c^{2}\right)\left(a+3b+5c\right)\Leftrightarrow9\left(a^{2}+3b^{2}+5c^{2}\right)\ge\left(a+3b+5c\right)^{2} \).
Aceasta ultima inegalitate este adevarata deoarece conform inegalitatii Cauchy-Buniakowski-Schwarz avem:
\( 9\left(a^{2}+3b^{2}+5c^{2}\right)=\left(1+3+5\right)\left(a^{2}+3b^{2}+5c^{2}\right)\ge\left(a+3b+5c\right)^{2}. \)