List of triangle inequalities

The parameters most commonly appearing in triangle inequalities are:

• the side lengths a, b, and c;
• the semiperimeter s = (a + b + c) / 2 (half the perimeter p);
• the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
• the values of trigonometric functions of the angles;
• the area T of the triangle;
• the medians m_a, m_b, and m_c of the sides (each being the length of the line segment from the midpoint of the side to the opposite vertex);
• the altitudes h_a, h_b, and h_c (each being the length of a segment perpendicular to one side and reaching from that side (or possibly the extension of that side) to the opposite vertex);
• the lengths of the internal angle bisectors t_a, t_b, and t_c (each being a segment from a vertex to the opposite side and bisecting the vertex's angle);
• the perpendicular bisectors p_a, p_b, and p_c of the sides (each being the length of a segment perpendicular to one side at its midpoint and reaching to one of the other sides);
• the lengths of line segments with an endpoint at an arbitrary point P in the plane (for example, the length of the segment from P to vertex A is denoted PA or AP);
• the inradius r (radius of the circle inscribed in the triangle, tangent to all three sides), the exradii r_a, r_b, and r_c (each being the radius of an excircle tangent to side a, b, or c respectively and tangent to the extensions of the other two sides), and the circumradius R (radius of the circle circumscribed around the triangle and passing through all three vertices).

The basic triangle inequality is

${\displaystyle a<b+c,\quad b<c+a,\quad c<a+b}$

or equivalently

${\displaystyle {\text{max}}(a,b,c)<s.}$

In addition,

${\displaystyle {\frac {3}{2}}\leq {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}<2,}$

where the value of the right side is the lowest possible bound,[1]^{:p. 259} approached asymptotically as certain classes of triangles approach the degenerate case of zero area. The left inequality, which holds for all positive a, b, c, is Nesbitt's inequality.

We have

${\displaystyle 3\left({\frac {a}{b}}+{\frac {b}{c}}+{\frac {c}{a}}\right)\geq 2\left({\frac {b}{a}}+{\frac {c}{b}}+{\frac {a}{c}}\right)+3.}$[2]^:p.250,#82

${\displaystyle abc\geq (a+b-c)(a-b+c)(-a+b+c).\quad }$[1]^{:p. 260}

${\displaystyle {\frac {1}{3}}\leq {\frac {a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}}<{\frac {1}{2}}.\quad }$[1]^{:p. 261}

${\displaystyle {\sqrt {a+b-c}}+{\sqrt {a-b+c}}+{\sqrt {-a+b+c}}\leq {\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}.}$[1]^{:p. 261}

${\displaystyle a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0.}$[1]^{:p. 261}

If angle C is obtuse (greater than 90°) then

${\displaystyle a^{2}+b^{2}<c^{2};}$

if C is acute (less than 90°) then

${\displaystyle a^{2}+b^{2}>c^{2}.}$

The in-between case of equality when C is a right angle is the Pythagorean theorem.

In general,[2]^:p.1,#74

${\displaystyle a^{2}+b^{2}>{\frac {c^{2}}{2}},}$

with equality approached in the limit only as the apex angle of an isosceles triangle approaches 180°.

If the centroid of the triangle is inside the triangle's incircle, then[3]^{:p. 153}

${\displaystyle a^{2}<4bc,\quad b^{2}<4ac,\quad c^{2}<4ab.}$

While all of the above inequalities are true because a, b, and c must follow the basic triangle inequality that the longest side is less than half the perimeter, the following relations hold for all positive a, b, and c:[1]^:p.267

${\displaystyle {\frac {3abc}{ab+bc+ca}}\leq {\sqrt[{3}]{abc}}\leq {\frac {a+b+c}{3}},}$

each holding with equality only when a = b = c. This says that in the non-equilateral case the harmonic mean of the sides is less than their geometric mean which in turn is less than their arithmetic mean.

${\displaystyle \cos A+\cos B+\cos C\leq {\frac {3}{2}}.}$ [1]^{:p. 286}

${\displaystyle (1-\cos A)(1-\cos B)(1-\cos C)\geq \cos A\cdot \cos B\cdot \cos C.}$[2]^:p.21,#836

${\displaystyle \cos ^{4}{\frac {A}{2}}+\cos ^{4}{\frac {B}{2}}+\cos ^{4}{\frac {C}{2}}\leq {\frac {s^{3}}{2abc}}}$

for semi-perimeter s, with equality only in the equilateral case.[2]^:p.13,#608

${\displaystyle a+b+c\geq 2{\sqrt {bc}}\cos A+2{\sqrt {ca}}\cos B+2{\sqrt {ab}}\cos C.}$ [4]^:Thm.1

${\displaystyle \sin A+\sin B+\sin C\leq {\frac {3{\sqrt {3}}}{2}}.}$ [1]^:p.286

${\displaystyle \sin ^{2}A+\sin ^{2}B+\sin ^{2}C\leq {\frac {9}{4}}.}$ [1]^{:p. 286}

${\displaystyle \sin A\cdot \sin B\cdot \sin C\leq \left({\frac {\sin A+\sin B+\sin C}{3}}\right)^{3}\leq \left(\sin {\frac {A+B+C}{3}}\right)^{3}=\sin ^{3}\left({\frac {\pi }{3}}\right)={\frac {3{\sqrt {3}}}{8}}.}$ [5]^{:p. 203}

${\displaystyle \sin A+\sin B\cdot \sin C\leq \varphi }$[2]^:p.149,#3297

where ${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}},}$ the golden ratio.

${\displaystyle \sin {\frac {A}{2}}\cdot \sin {\frac {B}{2}}\cdot \sin {\frac {C}{2}}\leq {\frac {1}{8}}.}$ [1]^{:p. 286}

${\displaystyle \tan ^{2}{\frac {A}{2}}+\tan ^{2}{\frac {B}{2}}+\tan ^{2}{\frac {C}{2}}\geq 1.}$ [1]^{:p. 286}

${\displaystyle \cot A+\cot B+\cot C\geq {\sqrt {3}}.}$ [6]

${\displaystyle \sin A\cdot \cos B+\sin B\cdot \cos C+\sin C\cdot \cos A\leq {\frac {3{\sqrt {3}}}{4}}.}$[2]^{:p.187,#309.2}

For circumradius R and inradius r we have

${\displaystyle \max \left(\sin {\frac {A}{2}},\sin {\frac {B}{2}},\sin {\frac {C}{2}}\right)\leq {\frac {1}{2}}\left(1+{\sqrt {1-{\frac {2r}{R}}}}\right),}$

with equality if and only if the triangle is isosceles with apex angle greater than or equal to 60°;[7]^{:Cor. 3} and

${\displaystyle \min \left(\sin {\frac {A}{2}},\sin {\frac {B}{2}},\sin {\frac {C}{2}}\right)\geq {\frac {1}{2}}\left(1-{\sqrt {1-{\frac {2r}{R}}}}\right),}$

with equality if and only if the triangle is isosceles with apex angle less than or equal to 60°.[7]^{:Cor. 3}

We also have

${\displaystyle {\frac {r}{R}}-{\sqrt {1-{\frac {2r}{R}}}}\leq \cos A\leq {\frac {r}{R}}+{\sqrt {1-{\frac {2r}{R}}}}}$

and likewise for angles B, C, with equality in the first part if the triangle is isosceles and the apex angle is at least 60° and equality in the second part if and only if the triangle is isosceles with apex angle no greater than 60°.[7]^{:Prop. 5}

Further, any two angle measures A and B opposite sides a and b respectively are related according to[1]^{:p. 264}

${\displaystyle A>B\quad {\text{if and only if}}\quad a>b,}$

which is related to the isosceles triangle theorem and its converse, which state that A = B if and only if a = b.

By Euclid's exterior angle theorem, any exterior angle of a triangle is greater than either of the interior angles at the opposite vertices:[1]^{:p. 261}

${\displaystyle 180{\text{°}}-A>\max(B,C).}$

If a point D is in the interior of triangle ABC, then

${\displaystyle \angle BDC>\angle A.}$[1]^{:p. 263}

For an acute triangle we have[2]^:p.26,#954

${\displaystyle \cos ^{2}A+\cos ^{2}B+\cos ^{2}C<1,}$

with the reverse inequality holding for an obtuse triangle.

Furthermore, for non-obtuse triangles we have[8]^{:Corollary 3}

${\displaystyle {\frac {2R+r}{R}}\leq {\sqrt {2}}\left(\cos \left({\frac {A-C}{2}}\right)+\cos \left({\frac {B}{2}}\right)\right)}$

with equality if and only if it is a right triangle with hypotenuse AC.

Weitzenböck's inequality is, in terms of area T,[1]^{:p. 290}

${\displaystyle a^{2}+b^{2}+c^{2}\geq 4{\sqrt {3}}\cdot T,}$

with equality only in the equilateral case. This is a corollary of the Hadwiger–Finsler inequality, which is

${\displaystyle a^{2}+b^{2}+c^{2}\geq (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+4{\sqrt {3}}\cdot T.}$

Also,

${\displaystyle ab+bc+ca\geq 4{\sqrt {3}}\cdot T}$[9]^{:p. 138}

and[2]^{:p.192,#340.3}[5]^{:p. 204}

${\displaystyle T\leq {\frac {abc}{2}}{\sqrt {\frac {a+b+c}{a^{3}+b^{3}+c^{3}+abc}}}\leq {\frac {1}{4}}{\sqrt[{6}]{\frac {3(a+b+c)^{3}(abc)^{4}}{a^{3}+b^{3}+c^{3}}}}\leq {\frac {\sqrt {3}}{4}}(abc)^{2/3}.}$

From the rightmost upper bound on T, using the arithmetic-geometric mean inequality, is obtained the isoperimetric inequality for triangles:

${\displaystyle T\leq {\frac {\sqrt {3}}{36}}(a+b+c)^{2}={\frac {\sqrt {3}}{9}}s^{2}}$ [5]^{:p. 203}

for semiperimeter s. This is sometimes stated in terms of perimeter p as

${\displaystyle p^{2}\geq 12{\sqrt {3}}\cdot T,}$

with equality for the equilateral triangle.[10] This is strengthened by

${\displaystyle T\leq {\frac {\sqrt {3}}{4}}(abc)^{2/3}.}$

Bonnesen's inequality also strengthens the isoperimetric inequality:

${\displaystyle \pi ^{2}(R-r)^{2}\leq (a+b+c)^{2}-4\pi T.}$

We also have

${\displaystyle {\frac {9abc}{a+b+c}}\geq 4{\sqrt {3}}\cdot T}$ [1]^{:p. 290}[9]^{:p. 138}

with equality only in the equilateral case;

${\displaystyle 38T^{2}\leq 2s^{4}-a^{4}-b^{4}-c^{4}}$[2]^:p.111,#2807

for semiperimeter s; and

${\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}<{\frac {s}{T}}.}$[2]^:p.88,#2188

Ono's inequality for acute triangles (those with all angles less than 90°) is

${\displaystyle 27(b^{2}+c^{2}-a^{2})^{2}(c^{2}+a^{2}-b^{2})^{2}(a^{2}+b^{2}-c^{2})^{2}\leq (4T)^{6}.}$

The area of the triangle can be compared to the area of the incircle:

${\displaystyle {\frac {\text{Area of incircle}}{\text{Area of triangle}}}\leq {\frac {\pi }{3{\sqrt {3}}}}}$

with equality only for the equilateral triangle.[11]

If an inner triangle is inscribed in a reference triangle so that the inner triangle's vertices partition the perimeter of the reference triangle into equal length segments, the ratio of their areas is bounded by[9]^{:p. 138}

${\displaystyle {\frac {\text{Area of inscribed triangle}}{\text{Area of reference triangle}}}\leq {\frac {1}{4}}.}$

Let the interior angle bisectors of A, B, and C meet the opposite sides at D, E, and F. Then[2]^:p.18,#762

${\displaystyle {\frac {3abc}{4(a^{3}+b^{3}+c^{3})}}\leq {\frac {{\text{Area of triangle}}\,DEF}{{\text{Area of triangle}}\,ABC}}\leq {\frac {1}{4}}.}$

A line through a triangle’s median splits the area such that the ratio of the smaller sub-area to the original triangle’s area is at least 4/9.[12]

The three medians ${\displaystyle m_{a},\,m_{b},\,m_{c}}$ of a triangle each connect a vertex with the midpoint of the opposite side, and the sum of their lengths satisfies[1]^{:p. 271}

${\displaystyle {\frac {3}{4}}(a+b+c)<m_{a}+m_{b}+m_{c}<a+b+c.}$

Moreover,[2]^:p.12,#589

${\displaystyle \left({\frac {m_{a}}{a}}\right)^{2}+\left({\frac {m_{b}}{b}}\right)^{2}+\left({\frac {m_{c}}{c}}\right)^{2}\geq {\frac {9}{4}},}$

with equality only in the equilateral case, and for inradius r,[2]^:p.22,#846

${\displaystyle {\frac {m_{a}m_{b}m_{c}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}}\geq r.}$

If we further denote the lengths of the medians extended to their intersections with the circumcircle as M_a , M_b , and M_c , then[2]^:p.16,#689

${\displaystyle {\frac {M_{a}}{m_{a}}}+{\frac {M_{b}}{m_{b}}}+{\frac {M_{c}}{m_{c}}}\geq 4.}$

The centroid G is the intersection of the medians. Let AG, BG, and CG meet the circumcircle at U, V, and W respectively. Then both[2]^:p.17#723

${\displaystyle GU+GV+GW\geq AG+BG+CG}$

and

${\displaystyle GU\cdot GV\cdot GW\geq AG\cdot BG\cdot CG;}$

in addition,[2]^:p.156,#S56

${\displaystyle \sin GBC+\sin GCA+\sin GAB\leq {\frac {3}{2}}.}$

For an acute triangle we have[2]^:p.26,#954

${\displaystyle m_{a}^{2}+m_{b}^{2}+m_{c}^{2}>6R^{2}}$

in terms of the circumradius R, while the opposite inequality holds for an obtuse triangle.

Denoting as IA, IB, IC the distances of the incenter from the vertices, the following holds:[2]^{:p.192,#339.3}

${\displaystyle {\frac {IA^{2}}{m_{a}^{2}}}+{\frac {IB^{2}}{m_{b}^{2}}}+{\frac {IC^{2}}{m_{c}^{2}}}\leq {\frac {3}{4}}.}$

The three medians of any triangle can form the sides of another triangle:[13]^{:p. 592}

${\displaystyle m_{a}<m_{b}+m_{c},\quad m_{b}<m_{c}+m_{a},\quad m_{c}<m_{a}+m_{b}.}$

Furthermore,[14]^{:Coro. 6}

${\displaystyle \max\{bm_{c}+cm_{b},\quad cm_{a}+am_{c},\quad am_{b}+bm_{a}\}\leq {\frac {a^{2}+b^{2}+c^{2}}{\sqrt {3}}}.}$

The altitudes h_a , etc. each connect a vertex to the opposite side and are perpendicular to that side. They satisfy both[1]^{:p. 274}

${\displaystyle h_{a}+h_{b}+h_{c}\leq {\frac {\sqrt {3}}{2}}(a+b+c)}$

and

${\displaystyle h_{a}^{2}+h_{b}^{2}+h_{c}^{2}\leq {\frac {3}{4}}(a^{2}+b^{2}+c^{2}).}$

In addition, if ${\displaystyle a\geq b\geq c,}$ then[2]^:222,#67

${\displaystyle a+h_{a}\geq b+h_{b}\geq c+h_{c}.}$

We also have[2]^:p.140,#3150

${\displaystyle {\frac {h_{a}^{2}}{(b^{2}+c^{2})}}\cdot {\frac {h_{b}^{2}}{(c^{2}+a^{2})}}\cdot {\frac {h_{c}^{2}}{(a^{2}+b^{2})}}\leq \left({\frac {3}{8}}\right)^{3}.}$

For internal angle bisectors t_a, t_b, t_c from vertices A, B, C and circumcenter R and incenter r, we have[2]^:p.125,#3005

${\displaystyle {\frac {h_{a}}{t_{a}}}+{\frac {h_{b}}{t_{b}}}+{\frac {h_{c}}{t_{c}}}\geq {\frac {R+4r}{R}}.}$

The reciprocals of the altitudes of any triangle can themselves form a triangle:[15]

${\displaystyle {\frac {1}{h_{a}}}<{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}},\quad {\frac {1}{h_{b}}}<{\frac {1}{h_{c}}}+{\frac {1}{h_{a}}},\quad {\frac {1}{h_{c}}}<{\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}.}$

The internal angle bisectors are segments in the interior of the triangle reaching from one vertex to the opposite side and bisecting the vertex angle into two equal angles. The angle bisectors t_a etc. satisfy

${\displaystyle t_{a}+t_{b}+t_{c}\leq {\frac {3}{2}}(a+b+c)}$

in terms of the sides, and

${\displaystyle h_{a}\leq t_{a}\leq m_{a}}$

in terms of the altitudes and medians, and likewise for t_b and t_c .[1]^{:pp. 271–3} Further,[2]^:p.224,#132

${\displaystyle {\sqrt {m_{a}}}+{\sqrt {m_{b}}}+{\sqrt {m_{c}}}\geq {\sqrt {t_{a}}}+{\sqrt {t_{b}}}+{\sqrt {t_{c}}}}$

in terms of the medians, and[2]^:p.125,#3005

${\displaystyle {\frac {h_{a}}{t_{a}}}+{\frac {h_{b}}{t_{b}}}+{\frac {h_{c}}{t_{c}}}\geq 1+{\frac {4r}{R}}}$

in terms of the altitudes, inradius r and circumradius R.

Let T_a , T_b , and T_c be the lengths of the angle bisectors extended to the circumcircle. Then[2]^:p.11,#535

${\displaystyle T_{a}T_{b}T_{c}\geq {\frac {8{\sqrt {3}}}{9}}abc,}$

with equality only in the equilateral case, and[2]^:p.14,#628

${\displaystyle T_{a}+T_{b}+T_{c}\leq 5R+2r}$

for circumradius R and inradius r, again with equality only in the equilateral case. In addition,.[2]^:p.20,#795

${\displaystyle T_{a}+T_{b}+T_{c}\geq {\frac {4}{3}}(t_{a}+t_{b}+t_{c}).}$

For incenter I (the intersection of the internal angle bisectors),[2]^:p.127,#3033

${\displaystyle 6r\leq AI+BI+CI\leq {\sqrt {12(R^{2}-Rr+r^{2})}}.}$

For midpoints L, M, N of the sides,[2]^:p.152,#J53

${\displaystyle IL^{2}+IM^{2}+IN^{2}\geq r(R+r).}$

For incenter I, centroid G, circumcenter O, nine-point center N, and orthocenter H, we have for non-equilateral triangles the distance inequalities[16]^:p.232

${\displaystyle IG<HG,}$

${\displaystyle IH<HG,}$

${\displaystyle IG<IO,}$

and

${\displaystyle IN<{\frac {1}{2}}IO;}$

and we have the angle inequality[16]^:p.233

${\displaystyle \angle IOH<{\frac {\pi }{6}}.}$

In addition,[16]^{:p.233,Lemma 3}

${\displaystyle IG<{\frac {1}{3}}v,}$

where v is the longest median.

Three triangles with vertex at the incenter, OIH, GIH, and OGI, are obtuse:[16]^:p.232

${\displaystyle \angle OIH}$ > ${\displaystyle \angle GIH}$ > 90° , ${\displaystyle \angle OGI}$ > 90°.

Since these triangles have the indicated obtuse angles, we have

${\displaystyle OI^{2}+IH^{2}<OH^{2},\quad GI^{2}+IH^{2}<GH^{2},\quad OG^{2}+GI^{2}<OI^{2},}$

and in fact the second of these is equivalent to a result stronger than the first, shown by Euler:[17][18]

${\displaystyle OI^{2}<OH^{2}-2\cdot IH^{2}<2\cdot OI^{2}.}$

The larger of two angles of a triangle has the shorter internal angle bisector:[19]^:p.72,#114

${\displaystyle {\text{If}}\quad A>B\quad {\text{then}}\quad t_{a}<t_{b}.}$

These inequalities deal with the lengths p_a etc. of the triangle-interior portions of the perpendicular bisectors of sides of the triangle. Denoting the sides so that ${\displaystyle a\geq b\geq c,}$ we have[20]

${\displaystyle p_{a}\geq p_{b}}$

and

${\displaystyle p_{c}\geq p_{b}.}$