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Pentru orice numere reale \( a\ge0 \),\( b\ge0 \),\( c\ge0 \) sa se demonstreze inegalitatea:

\( a^3+b^3+c^3\ge a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab} \).

\( (3,0,0)\succ (2,\frac{1}{2},\frac{1}{2}) \), by Muirhead it's true.

Another way: AM-GM: \( \frac{4a^3+b^3+c^3}{6}\ge\sqrt{a^4bc} \)

\( 4a^3+b^3+c^3\ge6a^2\sqrt{bc} \)
\( 4b^3+a^3+c^3\ge6b^2\sqrt{ac} \)
\( 4c^3+a^3+b^3\ge6c^2\sqrt{ab} \)
Sum, and that's it!

Din inegalitatea mediilor avem: \( \sum a^{2}\sqrt{bc}\le\sum a^{2}\cdot\frac{b+c}{2}=\frac{1}{2}\sum a^{2}\left(b+c\right) \).

Problema revine la \( \sum a^{2}\sqrt{bc}\le\sum a^{3}\Longleftrightarrow2\sum a^{3}\ge\sum a^{2}\left(b+c\right)=\sum ab\left(a+b\right)\left(*\right) \).

Dar \( a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\Longleftrightarrow\sum a^{3}+b^{3}\ge\sum ab\left(a+b\right)\Longleftrightarrow\left(*\right) \)

Claudiu Mindrila wrote:

Problema revine la \( \sum a^{2}\sqrt{bc}\le\sum a^{3}\Longleftrightarrow2\sum a^{3}\ge\sum a^{2}\left(b+c\right)=\sum ab\left(a+b\right)\left(*\right) \).

Atentie la scriere:

Corect e \( \sum a^{2}\sqrt{bc}\le\sum a^{3}\Longleftarrow2\sum a^{3}\ge\sum a^{2}\left(b+c\right)=\sum ab\left(a+b\right)\left(*\right) \).

-impartind totul prin \( abc \) vom avea de demonstrat:
\( \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}\ge \frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ac}}+\frac{c}{\sqrt{ab}}. \)
-notam \( \frac{a}{\sqrt{bc}}=x,\frac{b}{\sqrt{ac}}=y,\frac{c}{\sqrt{ab}}=z\Longrightarrow xyz=1 \).
-ramane de demonstrat ca \( x^2+y^2+z^2\ge x+y+z \), dar \( x+y+z\ge 3\sqrt[3]{xyz}\Longrightarrow x+y+z\ge 3 \).
\( x^2+y^2+z^2\ge \frac{(x+y+z)^2}{3} \) si cum \( x+y+z\ge 3 \) rezulta concluzia.