Inegalitate cu logaritmi conditionata

[Radu Titiu]

Daca \( a,b,c>0 \) si \( a+b+c=1 \) aratati ca:

\( \frac{a^2}{\log_a(a^2+b^2+c^2)}+\frac{b^2}{\log_b(a^2+b^2+c^2)}+\frac{c^2}{\log_c(a^2+b^2+c^2)}\geq \log_{abc}(a^2+b^2+c^2) \)

[Marius Mainea]

Avem \( (\sum {a^2\log _{a^2+b^2+c^2}^a})(\sum {\log_{a^2+b^2+c^2}^a})\geq1 \)

Aplicand CBS \( LHS\geq(\sum {a\log_{a^2+b^2+c^2}^a})^2\geq (\log_{a^2+b^2+c^2}^{a^2+b^2+c^2})^2=1 \)


deoarece functia \( f(x)=\log_{a^2+b^2+c^2}^x \) este convexa, si aplicam inegalitatea lui Jensen.

\( f(ax+by+cz)\leq af(x)+bf(y)+cf(z) \) pentru a+b+c=1 cu a, b, c nenegative.