# Geometric Inequalities: Introduction

When not stated otherwise, lengths of the sides of a triangle \( ABC \) are labeled by
\( a=BC \), \( b=CA \), \( c=AB \). The angles are denoted \( \alpha=\angle A=\angle BAC \), \( \beta=\angle B=\angle ABC \), \( \gamma=\angle C=\angle ACB \). The midpoints of the sides \( BC \), \( CA \), \( AB \) are denoted by \( A_1 \), \( B_1 \), \( C_1 \), and the feet of the altitudes from \( A \), \( B \), \( C \) to the opposite sides by \( A^{\prime} \), \( B^{\prime} \), and \( C^{\prime} \). We will frequently denote the points where the internal bisectors intersect the sides of the triangle by \( A^{\prime\prime} \), \( B^{\prime\prime} \), \( C^{\prime\prime} \).
The lengths of the medians \( AA_1 \), \( BB_1 \), \( CC_1 \) are denoted by \( m_a \), \( m_b \), \( m_c \); the lengths of the altitudes \( AA^{\prime} \), \( BB^{\prime} \), \( CC^{\prime} \) by \( h_a \), \( h_b \), \( h_c \); and the lengths of the segments of internal bisectors by \( l_a=AA^{\prime\prime} \), \( l_b=BB^{\prime\prime} \), and \( l_c=CC^{\prime\prime} \). The semi-perimeter of the triangle will be denoted by \( p \) (i.e.
\( p=(a+b+c)/2 \)). The circumradius of the triangle will be denoted by \( R \) and the inradius by \( r \). The radii of the three circles tangent to one side and the extensions of the other two sides (called the *excircles*) will be denoted by \( r_a \), \( r_b \), \( r_c \). \( S \) and \( S_{ABC} \) will denote the area of the triangle \( ABC \).

** Theorem 1 (Triangle inequality) **

If \( ABC \) is a triangle then the following statements hold:

**(a)**
If \( a \), \( b \), \( c \) are the lengths of the sides, then \( a< b+c \), \( b< c+a \), \( c< a+b \). Conversely, if \( a \), \( b \), \( c \) are positive real numbers each of which is smaller than the sum of the other two, then there exists a triangle whose side lengths are \( a \), \( b \), and \( c \).
**(b)** \( AB< BC \) if and only if \( \angle ACB< \angle BAC \).

** Problem 1 **

Prove that for arbitrary triangle the following inequalities hold: \[ p< m_a+m_b+m_c < 2p.\]

Let \( ABC \) be the given triangle, \( A_1 \), \( B_1 \), \( C_1 \) the midpoints of \( BC \), \( CA \), \( AB \). Let \( E \) be the point such that \( ABEC \) is a parallelogram. Then \( AE \) passes through \( A_1 \) and \( AE=2m_a \). From the triangle \( ABE \) we get
\( AE< AB+BE=c+b \) since \( BE=AC=b \). Therefore \( 2m_a< b+c \). Writing down two analogous inequalities \( 2m_b< c+a \), \( 2m_c< a+b \) which are proved in the same way and summing them up we get
\( 2m_a+2m_b+2m_c< 2a+2b+2c \) and one half of the problem is solved.
For the other half notice that from \( \triangle AA_1B \) we conclude \( c=AB< AA_1+BA_1=m_a+\frac a2 \). Similarly from the triangle \( AA_1C \) we get \( b< m_a+\frac a2 \) and summing the last two inequalities yields \( b+c< a+2m_a \). Adding this with the two analogous inequalities gives \( 2a+2b+2c< a+b+c+2m_a+2m_b+2m_c \). This implies that \( a+b+c< 2(m_a+m_b+m_c) \) or equivalently \( p< m_a+m_b+m_c \). It is also obvious now that both inequalities are always strict.

** Problem 2 **

Prove that for every triangle the sum of its medians is greater than \( 3/4 \) of the sum of its sides.

Let \( ABC \) be the given triangle, \( A_1 \), \( B_1 \), \( C_1 \) the midpoints of \( BC \), \( CA \), \( AB \), \( G \) its centroid (the intersection of its medians \( AA_1 \), \( BB_1 \), and \( CC_1 \)). We know that \( GA=\frac23 AA_1 \), \( GB=\frac23 BB_1 \), and \( GC=\frac23 CC_1 \). From \( a=BC< BG+GC=\frac23(BB_1+CC_1)=\frac23(m_b+m_c) \) and two analogous inequalities we get
\( a+b+c< \frac43(m_a+m_b+m_c) \). Then we have \( m_a+m_b+m_c> \frac34(a+b+c) \).

** Theorem 2 (Ptolemy) **

For
any four points \( A \), \( B \), \( C \), \( D \), in the plane \[ AC\cdot BD\leq AB\cdot
CD+AD\cdot BC.\]
The equality holds if and only if \( ABCD \) is cyclic with diagonals \( AC \) and \( BD \); or if \( A \), \( B \), \( C \), \( D \) are collinear and exactly one of \( B \), \( D \) is between \( A \) and \( C \).

Let \( M \) be the point in the plane such that \( \triangle CMB \) and \( \triangle CDA \) are similar and equally oriented. Then \( \frac{CM}{BC}=\frac{CD}{AC} \) and
\( \angle BCM=\angle ACD \). Hence \( \angle DCM=\angle ACB \) which implies \( \triangle DCM\sim\triangle ACB \). Therefore \( BM=\frac{BC\cdot AD}{AC} \) and \( MD=\frac{CD\cdot AB}{AC} \) hence \( BD\leq BM+MD=\frac{BC\cdot AD+CD\cdot AB}{AC} \). This proves the inequality.
If the equality holds, then \( B \), \( M \), \( D \) must be collinear. This in turn implies
\( \angle CBD=\angle CAD \) hence \( ABCD \) is cyclic or the points \( A \), \( B \), \( C \), \( D \) are collinear. In the last case it is easy to see that the stated condition must be satisfied.

** Theorem 3 (Parallelogram Inequality) **

For every four points \( A \), \( B \), \( C \), \( D \) in the space we have
\[ AB^2+BC^2+CD^2+DA^2\geq AC^2+BD^2.\] The equality holds if an only if \( ABCD \) is a parallelogram (or degenerated parallelogram).

Consider this problem in the coordinate system. Assume that \( A(x_1,y_1,z_1) \), \( B(x_2,y_2,z_2) \), \( C(x_3,y_3,z_3) \), and \( D(x_4,y_4,z_4) \). The required inequality will follow if we prove \[ (x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_4)^2+(x_4-x_1)^2\geq
(x_1-x_3)^2+(x_2-x_4)^2\] and the analogous inequalities when \( x \) is replaced by \( y \) and \( z \). The last inequality is equivalent to
\( 0\leq x_1^2+x_2^2+x_3^2+x_4^2-2x_1x_2-2x_2x_3-2x_3x_4-2x_4x_1+2x_1x_3+2x_2x_4=
(x_1-x_2+x_3-x_4)^2 \) which is true.
The equality holds if and only if \( x_1+x_3=x_2+x_4 \), \( y_1+y_3=y_2+y_4 \), and \( z_1+z_3=z_2+z_4 \), i.e. if and only if \( ABCD \) is a parallelogram.

** Problem 3 **

Let \( ABC \) be an acute-angled triangle. Using a straight-edge and compass construct a point \( M \) inside the triangle \( ABC \) for which the sum \( MA+MB+MC \) is minimal.

Let \( X \) be any point in the interior of \( \triangle ABC \). Let \( Q \) be the point outside the triangle for which \( BAQ \) is equilateral. Point \( Q \) is the image of \( B \) under the rotation around \( A \) for the angle \( 60^{\circ} \). Let \( Y \) be the image of \( X \) under this rotation. Then \( \triangle XAY \) is equilateral, and \( \triangle QAY\equiv
\triangle BAX \). Therefore \( QY=BX \) and \( YX=AX \). Hence \( AX+BX+CX=QY+YX+CX\geq QC \).
Now it suffices to construct a point \( M \) for which \( MA+MB+MC=QC \).
This can be done because we can construct points \( M \) and \( M^{\prime} \) on \( QC \)
for which \( M^{\prime}AM \) is equilateral (just construct two lines from \( A \) that are at an angle \( 60^{\circ} \) to \( QC \)).

*Remark.* Point \( M \) obtained in the previous example is called the *Toricelli point* of the triangle \( ABC \). From the previous proof it follows that \( M \) is the intersection of \( AQ_A \), \( BQ_B \), and \( CQ_C \) where \( Q_A \), \( Q_B \), and \( Q_C \) are the points in the exterior of \( \triangle ABC \) for which \( \triangle BAQ_C \), \( \triangle ACQ_B \), and \( \triangle BQ_AC \) are equilateral.

** Problem 4 **

Prove that \( h_a\leq l_a\leq m_a \).

The inequality \( h_a\leq l_a \) is obvious because the altitude is the shortest segment connecting \( A \) with some point on \( BC \).
Let us now concentrate on \( l_a\leq m_a \).
If \( AB=AC \) then \( l_a=m_a \) and there is nothing to prove. Let us assume assume that \( AC> AB \).
Let \( M \) be the point symmetric to \( A \) with respect to \( A_1 \). Then \( BMCA \) is a parallelogram with \( BM=AC \). From \( BM=AC> AB \) we conclude that \( \angle BMA< \angle BAM \) and therefore \( \angle A_1AC< \angle BA A_1 \). This implies that \( AA^{\prime\prime} \) belongs to the interior of the angle \( \angle BAA_1 \). Furthermore, \( \angle AA^{\prime\prime}C \) is obtuse because
\( \angle AA^{\prime\prime}C+\angle BA^{\prime\prime}A=180^{\circ} \) and \( \angle AA^{\prime\prime}C=\angle A^{\prime\prime}BA+\angle BAA^{\prime\prime}
=\angle A^{\prime\prime}BA+\alpha/2> A^{\prime\prime}CA+\alpha/2=\angle BA^{\prime\prime}A \). Thus \( AA_1 \) is the longest side of \( \triangle AA^{\prime\prime}A_1 \) hence \( AA_1> AA^{\prime\prime} \) or \( m_a> l_a \).