SA+SB+SC>3 if MA>1, MB>1,MC>1 where M centroid of ABC in pyramid SABC

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SA+SB+SC>3 if MA>1, MB>1,MC>1 where M centroid of ABC in pyramid SABC

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#1 h Aug 5, 2020, 7:47 PM

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Let $SABC$ be such a triangular pyramid that for the point $M$ , its median intersection of the face $ABC$ , hold the inequalities $MA>1, MB>1$ and $MC>1$ . Prove that $SA+SB+SC>3$

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#2 h Aug 8, 2020, 8:15 PM

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Let $SH$ be the altitude of the pyramid and let $T$ be the Torricelli point of $ABC$ . Note that
$SA+SB+SC>HA+HB+HC\ge TA+TB+TC$ thus it suffices to prove the inequality
$TA+TB+TC>3.$ Firstly consider the case when any angle of $\Delta ABC$ is less than or equal to $120^\circ$ . Denote $TA=x, TB=y, TC=z$ and let $z=\min\{x, y, z\}$ . Then $AB=\sqrt{x^2+xy+y^2}$ and so on. by the given condition the length of the $C$ -median is greater than $3/2:$
$\sqrt{\frac{2(x^2+xz+z^2)+2(y^2+yz+z^2)-(x^2+xy+y^2)}{4}}>\frac{3}{2}$ $\iff$
$(x+y+z)^2+3z^2-3xy>9.$ Summing up this with the inequality
$3xy-3z^2\ge 0$ we get
$x+y+z>3.$ Now consider the case when one of the angles of $\Delta ABC$ is greater than $120^\circ$ . WLOG it is the angle $C$ . Then $T\equiv C$ . Denote $AB=c, BC=a, CA=b$ . We need to prove that
$a+b>3.$ Again note that the $C$ -median is greater than $3/2:$
$\sqrt{\frac{2a^2+2b^2-c^2}{4}}>\frac{3}{2}.$ Hence
$(a+b)^2>a^2+b^2>(a^2+b^2)+(a^2+b^2-c^2)>9$ and thus
$a+b>3$ as desired.

This post has been edited 1 time. Last edited by kiyoras_2001, Aug 9, 2020, 10:12 AM

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