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Fie \( a,b,c \) trei numere reale pozitive cu suma \( 1 \) astfel incat \( 3a-b \geq0 \) , \( 3b-c \geq0 \) si \( 3c-a \geq0 \) . Aratati ca \( \sum \frac{a^3}{3a-b} \geq \frac{1}{6} \) .

Mihai Opincariu, G.M.B. 11/1999

Folosim inegalitatea:

\( \underline{\overline{\left\|\ \frac{a^3}{x}\ +\ \frac{b^3}{y}\ +\ \frac{c^3}{z}\ \ge\ \frac{(a+b+c)^3}{3(x+y+z)}\ ,\ (\forall)\ a\ ,\ b\ ,\ c\ ,\ x\ ,\ y\ ,\ z\ >\ 0\ (*)\ \right\|}} \)

Prin urmare, \( LHS\ \ge\ \frac{(a+b+c)^3}{3(3a-b+3b-c+3c-a)}=\frac{(a+b+c)^3}{6(a+b+c)}=\frac 16 \)

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Demonstratia inegalitatii \( (*) \)

Aplicand inegalitatea lui Cauchy-Schwarz avem: \( \left\(\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\right\)(x+y+z)\ge \left\(a^{\frac 32}+b^{\frac 32}+c^{\frac 32}\right)^2\ (1) \) .

Dar din inegalitatea mediilor generalizata avem: \( \left\(\frac{a^{\frac 32}+b^{\frac 32}+c^{\frac 32}}{3}\right\)^{\frac 23}\ \ge\ \frac{a+b+c}{3}\ \Longleftrightarrow\ \left\(a^{\frac 32}+b^{\frac 32}+c^{\frac 32}\right\)^2\ \ge\ \frac{(a+b+c)^3}{3}\ (2) \) .

Din \( (1) \) si \( (2) \) rezulta in mod evident inegalitatea \( (*) \) .

O solutie la nivel de clasa a VIII-a ar arata astfel :
\( 4(3a-b)(a+b) \leq(3a-b+a+b)^2=16a^2 \) de unde obtinem ca \( \frac{a^3}{3a-b} \geq \frac{a^2+ab}{4} \) si analoagele care insumate duc la : \( \sum \frac{a^3}{3a-b} \geq \frac{ \sum (a^2+bc)}{4} \) si folosim in continuare faptul ca \( \sum (a^2+bc) \geq \frac{2(a+b+c)^2}{3} \).