Fie \( a, b, c\in \mathb{R} \) astfel incat \( a^2+b^2+c^2=x>0 \). Sa se arate ca:
\( a^3+b^3+c^3-3abc\le x\sqrt{x} \).
Marcel Chirita, Curtea de Arges
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Notam \( y=ab+bc+ca \) . Atunci din \( (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ \Longrightarrow\ a+b+c=\sqrt{x+2y} \) .
Folosim identitatea cunoscuta \( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \) , care, cu notatiile facute devine :
\( a^3+b^3+c^3-3abc=\sqrt{x+2y}\cdot (x-y)=\sqrt{x+2y}\cdot\sqrt{x-y}\cdot\sqrt{x-y} \) , intrucat \( x\ge y \) .
Dar \( \sqrt{x+2y}\cdot\sqrt{x-y}\cdot\sqrt{x-y}\ \le^{\small{AM-GM}}\ \left\(\frac{\sqrt{x+2y}+\sqrt{x-y}+\sqrt{x-y}}{3}\right\)^3\ \le^{\small{CBS}}\ \left\(\frac{\sqrt{3[(x+2y)+(x-y)+(x-y)]}}{3}\right\)^3=x\sqrt{x} \) .
Asadar , \( a^3+b^3+c^3-3abc\le x\sqrt x \) .
Folosim identitatea cunoscuta \( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \) , care, cu notatiile facute devine :
\( a^3+b^3+c^3-3abc=\sqrt{x+2y}\cdot (x-y)=\sqrt{x+2y}\cdot\sqrt{x-y}\cdot\sqrt{x-y} \) , intrucat \( x\ge y \) .
Dar \( \sqrt{x+2y}\cdot\sqrt{x-y}\cdot\sqrt{x-y}\ \le^{\small{AM-GM}}\ \left\(\frac{\sqrt{x+2y}+\sqrt{x-y}+\sqrt{x-y}}{3}\right\)^3\ \le^{\small{CBS}}\ \left\(\frac{\sqrt{3[(x+2y)+(x-y)+(x-y)]}}{3}\right\)^3=x\sqrt{x} \) .
Asadar , \( a^3+b^3+c^3-3abc\le x\sqrt x \) .