Inegalitate ,,draguta''

[Marius Mainea]

Daca a,b,c sunt pozitive atunci :
\( \frac{a^3}{(b+c)^2}+\frac{b^3}{(c+a)^2}+\frac{c^3}{(a+b)^2}\ge\frac{a+b+c}{4} \)

Gh. Stoica, Revista Arhimede 1-6/2008

[Claudiu Mindrila]

Fara a leza generalitatea, putem presupune ca \( a \le b\le c \). Atunci, aplicand inegalitatea rearanjamentelor, avem:

\( \left|\begin{array}{c}
a^{3}\le b^{3}\le c^{3}\\
\frac{1}{\left(b+c\right)^{2}}\le\frac{1}{\left(c+a\right)^{2}}\le\frac{1}{\left(a+b\right)^{2}}\end{array}\right|\Longrightarrow\left|\begin{array}{c}
\sum\frac{a^{3}}{\left(c+a\right)^{2}}\le\sum\frac{a^{3}}{\left(b+c\right)^{2}}\\
\sum\frac{c^{3}}{\left(c+a\right)^{2}}\le\sum\frac{a^{3}}{\left(b+c\right)^{2}}\end{array}\right|\oplus\Longrightarrow\sum\frac{a^{3}}{\left(b+c\right)^{2}}\ge\frac{1}{2}\cdot\sum\frac{a^{3}+c^{3}}{\left(a+c\right)^{2}} \).

Acum, aplicand inegalitatea lui Cebisev, avem:
\( \frac{1}{2}\cdot\sum\frac{a^{3}+c^{3}}{\left(a+c\right)^{2}}\ge\frac{1}{2}\cdot\sum\frac{\left(a+c\right)\left(a^{2}+c^{2}\right)}{2\left(a+c\right)^{2}}\ge\frac{1}{2}\cdot\sum\frac{\left(a+c\right)\left(a+c\right)^{2}}{4\left(a+c\right)^{2}}=\frac{1}{2}\cdot\sum\frac{a+c}{4}=\frac{\sum\frac{a}{2}}{2}=\sum\frac{a}{4} \), ceea ce trebuia aratat.

[zeta]

\( \sum{\frac{1}{2}(\frac{a}{b+c})^3\frac{b+c}{a+b+c}}\geq\ (\sum{\frac{1}{2}\frac{a(b+c)}{(b+c)(a+b+c)}})^3=\frac{1}{8} \), de unde rezulta ineg din enunt.

[zeta]

Mai sus am aplicat inegalitatea lui Jensen.

[Virgil Nicula]

Mi-a placut mult, Claudiu ! Frumos demonstrat si limpede redactat.
Inca odata, felicitari ! Cresti de la o zi la alta.