Inegalitate polinomiala 4

Marius Mainea · Post by **Marius Mainea** » Tue Dec 01, 2009 7:53 pm

Daca a, b, c, d sunt numere reale cu \( a+b+c+d=0 \), atunci \( a^4+b^4+c^4+d^4+28abcd\ge 0. \)

Marius Mainea · Post by **Marius Mainea** » Sat Dec 05, 2009 8:26 pm

Indicatie:

Putem presupune fara a pierde generalitatea ca \( a,b,c\ge 0 \) si \( d\le 0 \).

Atunci \( d=-a-b-c \) si inegalitatea devine \( a^4+b^4+c^4+(a+b+c)^4-28abc(a+b+c)\ge 0 \) cu a, b, c nenegative.

Claudiu Mindrila · Post by **Claudiu Mindrila** » Sat Dec 05, 2009 9:04 pm

Problema revine la \( \left(a^{4}+b^{4}+c^{4}-abc(a+b+c)\right)+\left(a+b+c\right)\left(\left(a+b+c\right)^{3}-27abc\right)\ge0 \), relatie adevarata tinand cont ca ambele cantitati din paranteze sunt pozitive (exercitiu).

Marius Mainea · Post by **Marius Mainea** » Sat Dec 05, 2009 9:59 pm

Poti sa detaliezi?

Claudiu Mindrila · Post by **Claudiu Mindrila** » Sat Dec 05, 2009 10:04 pm

\( \sum a^{4}\ge\sum a^{2}b^{2}\ge\sum ab\cdot bc=abc\left(\sum a\right) \) si \( a+b+c\ge3\sqrt[3]{abc}\Longrightarrow\left(a+b+c\right)^{3}\ge27abc \)