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Aratati ca pentru orice \( a,b,c > 0 \) avem:
\( \sqrt{(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)} \geq abc + \sqrt[3] {(a^3+abc)(b^3+abc)(c^3+abc)} \)

Se pare ca nu le prea plac celor de-a 9-a inegalitatile.

\( \sqrt {({\sum_{ciclic}^{} {a^2b}})({\sum_{ciclic}^{} {ab^2})}}= \frac {1}{2} {\sqrt {[b(a^2+bc)+c(b^2+ca)+a(c^2+ab)][c(a^2+bc)+a(b^2+ca)+b(c^2+ab)]} \)

\( \frac {Cauchy}{\geq} \) \( {\frac {1}{2}}({{\sqrt {bc}}(a^2+bc)+{\sqrt {ca}}(b^2+ca)+{\sqrt {ab}}(c^2+ab)}) \)

\( \frac {m_a-m_g}{\geq} \) \( {\frac {3}{2}}{\sqrt[3] {abc(a^2+bc)(b^2+ca)(c^2+ab)}}= {\frac {1}{2}} \sqrt[3] {(a^3+abc)(b^3+abc)(c^3+abc)} + \sqrt[3] {(a^3+abc)(b^3+abc)(c^3+abc)} \)

\( \frac {m_a-m_g}{\geq} \) \( abc + \sqrt[3] {(a^3+abc)(b^3+abc)(c^3+abc)} \).

Sau daca vreti o rezolvare mai simpla:
Impartim prin \( abc \) si notam \( x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a} \).
\( s=x+y+z, q=xy+yz+zx, p=xyz=1, t=\sqrt{sq} \).
Inegalitatea se transforma treptat in:
\( \sqrt{(x+y+z)(xy+yz+zx)}-1\geq\sqrt[3]{(x+y)(y+z)(z+x)}\Leftrightarrow \)
\( \sqrt{sq}-1\geq\sqrt[3]{sq-1} \Leftrightarrow \)
\( t(t-1)(t-3)\geq 0 \), care e adevarata din cauza ca \( t\geq3 \) din \( m_a\geq m_g \).
Egalitate cand \( x=y=z\rightarrow a=b=c \).