Dan Barbilian 2008

[andy crisan]

Sa se arate ca in orice triunghi are loc relatia.
\( 2^a+2^b+2^c>2^m^a+2^m^b+2^m^c \), notatiile fiind cele cunoscute.

[mihai++]

avem \( \sum2^a=\sum\frac{2^a+2^b}{2}\geq\sum2^{\frac{a+b}{2}}>\sum2^{m_c} \), deoarece:
\( m_c^2=\frac{2(a^2+b^2)-c^2}{4}\leq\frac{2(a^2+b^2)-(a-b)^2}{4}=(\frac{a+b}{2})^2 \).