Inegalitate in numere reale pozitive

[Mateescu Constantin]

Demonstrati ca pentru orice numere reale pozitive \( a,\ b,\ c \) avem inegalitatea:

\( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq 3\sqrt[4]{\frac{a^{4}+b^{4}+c^{4}}{3}} \).

[alex2008]

Se poate demonstra inegalitatea mai tare :

\(
\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 3\sqrt[6]{\frac{a^6+b^6+c^6}{3}}
\)

A se vedea aici

Aici este generalizarea inca nerezolvata.

[alex2008]

Pentru inegalitatea initiala :

\( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\sqrt[4]{\frac{a^{4}+b^{4}+c^{4}}{3}} \)

Aplicam Holder :

\( \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2\left(a^2b^2+b^2c^2+c^2a^2\right) \ge (a^2+b^2+c^2)^3 \)

Notam : \( a^2=x;b^2=y;c^2=z \)

Deci ar mai trebui sa demonstram ca :

\( \frac{(x+y+z)^3}{xy+yz+zx} \ge 9\sqrt{\frac{x^2+y^2+z^2}{3}}
\Leftrightarrow \)\( (x+y+z)^3 \ge 3\sqrt{3}(xy+yz+zx)\sqrt{x^2+y^2+z^2} \)

Avem ca :
\( (x+y+z)^3 =((x+y+z)^2)^{\frac{3}{2}}=(\sum x^2+2\sum xy)^{\frac{3}{2}}
\ge \)\( \left(3\sqrt[3]{\left(\sum x^2\right)\left(\sum xy\right)^2}\right)^{\frac{3}{2}}=3\sqrt{3}(xy+yz+zx)\sqrt{x^2+y^2+z^2} \)