Y by Adventure10
Prove that in any acute or right-angled triangle
with circumradius
, inradius
, semiperimeter
and medians of lengths
,
,
, the following inequality holds:
![\[\boxed{m_a + m_b + m_c \ge \sqrt{\frac {5s^2 + 12Rr + 3r^2}2}}\ .\]](//latex.artofproblemsolving.com/3/5/4/35419a162729e890c0c83117e634a9c6d1cb4bd5.png)
Remark. The proposed inequality can be written in the following equivalent forms:
![\[
\boxed{
\begin{array}{ll}
1 \blacktriangleright & \sum \left(m_a + m_b\right)\left(m_a + m_c\right) \ge 3\sum bc \\\\
2 \blacktriangleright & \sum \left(m_b + m_c\right)^2 \ge 4s^2 \\\\
3 \blacktriangleright & \sum \left(b - c\right)^2 \ge \sum \left(m_b - m_c\right)^2 \\\\
4 \blacktriangleright & \Big(\sum m_a\Big)^2 \ge \frac 14\Big(\sum a^2 + 8\sum bc\Big)
\end{array}
}\ .
\]](//latex.artofproblemsolving.com/3/e/0/3e00d82e480057d8ff7979706bcdcb1d7677dc85.png)







![\[\boxed{m_a + m_b + m_c \ge \sqrt{\frac {5s^2 + 12Rr + 3r^2}2}}\ .\]](http://latex.artofproblemsolving.com/3/5/4/35419a162729e890c0c83117e634a9c6d1cb4bd5.png)
Remark. The proposed inequality can be written in the following equivalent forms:
![\[
\boxed{
\begin{array}{ll}
1 \blacktriangleright & \sum \left(m_a + m_b\right)\left(m_a + m_c\right) \ge 3\sum bc \\\\
2 \blacktriangleright & \sum \left(m_b + m_c\right)^2 \ge 4s^2 \\\\
3 \blacktriangleright & \sum \left(b - c\right)^2 \ge \sum \left(m_b - m_c\right)^2 \\\\
4 \blacktriangleright & \Big(\sum m_a\Big)^2 \ge \frac 14\Big(\sum a^2 + 8\sum bc\Big)
\end{array}
}\ .
\]](http://latex.artofproblemsolving.com/3/e/0/3e00d82e480057d8ff7979706bcdcb1d7677dc85.png)