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Inequality in a triangle

Given a triangle ABC let m_a , m_b , m_c denote the median points of the sides a , b, c respectively. Prove that

    \[\left ( m_a + m_b + m_c \right )^2 + 4 r \left ( R - 2r \right ) \leq \left ( 4 R + r \right )^2\]

where R denotes the the circumradius and r the inradius respectively.

(Adil Abdullayev / RMM)

Solution [Soumava Chakraborty]

We have successively

    \begin{align*} \left ( \sum m_a \right )^2 &=\sum m_a^2 + 2 \sum m_a m_b \\ &=\frac{3}{4} \sum a^2 + 2 \sum m_a m_b \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{m_am_b \leq \frac{2c^2+ab}{4}}{\leq } \frac{3}{4} \sum a^2 + \frac{1}{2} \sum \left ( 2c^2 + ab \right )\\ &= \frac{3}{4} \sum a^2 + \sum a^2 + \frac{1}{2} \sum ab\\ &= \frac{7}{4} \sum a^2 + \frac{1}{2} \sum ab \\ &= \frac{1}{4} \left [ \sum a^2 + 2 \sum ab + 6 \sum a^2 \right ]\\ &= \frac{4s^2 + 12 \left ( s^2-4Rr -r^2 \right )}{4} \\ &=4s^2 -12 Rr -3r^2 \\ &\!\!\!\!\!\!\!\!\!\overset{\text{Gerretsen}}{\leq } 4 \left ( 4R^2 + 4Rr + 3r^2 \right ) -12 Rr -3r^2 \\ &=16R^2 + 8Rr + r^2 -4Rr + 8r^2 \\ &=\left ( 4R+r \right )^2 - 4r \left ( R - 2r \right ) \end{align*}

In all the above s denotes the semiperimeter of the triangle. More on Gerretsen’s inequality can be found at this link .

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