after many efforts, I found a good proof for the

inequality IA+IB+IC <= GA+GB+GC (1)

based on an interesting property of X(177).

Let A'B'C' be the cevian triangle of the incenter I,

let A1B1C1 be the intouch triangle of ABC and

let A2B2C2 be the triangle with vertices on the

incircle and the segments IA, IB, IC that are the

mid-arc points of triangle A'B'C'.

The line A1A2 is a internal bisector of triangle

A'B'C'

and is perpendicular to B2C2.

Hence A1A2, B1B2, C1C2 pass through the

incenter Q of A'B'C' that is the orthocenter of

A2B2C2.

This point Q is the point X(177) in ETC.

If c < b and M1 is the mid-point of BC then

BA1 = s-b < BA' = ac/(b+c) < BM1 = a/2.

Hence the point Q is on the other side of the centroid

G

relative to the line AA'.

The lines AA', BB', CC' devide the triangle ABC in 6

triangles

Tba = IBA', Tcb = ICB', Tac = IAC'

Tca = IA'C, Tab= IB'A, Tbc = IC'B

If the points G, I coincide then (1) is equality.

If G is on a bisector of ABC then Q is on the same

bisector

and the angle GIQ = pi.

If e.g. G is inside triangle Tbc then Q is inside

triangle Tcb

and since angle BIC is obtuse then also angle GIQ

is also obtuse and hence always

the dot product of vectors IG, IQ

is negative or zero i.e

IG.IQ <= 0 (2)

Let i, j, k be the unit vectors along IA, IB, IC

and r be the inradius of ABC.

The points Q, I are the orthocenter and circumcenter

of A2B2C2

and for vectors Sylvester's theorem gives

IA2 + IB2 + IC2 = IQ or

r(i + j + k) = IQ or

i + j + k = (1/r)IQ

We have

IA + IB + IC = IA.i + IB.j + IC.k

= (IG +GA).i + (IG + GB).j + (IG + GC).k

= IG.(i + j + k) + GA.i + GB.j + GC.k

= (1/r)IG.IQ + GA.i + GB.j + GC.k

<= GA.i + GB.j + GC.k due to (2)

and hence

IA + IB + IC <= | GA.i + GB.j + GC.k |

<= | GA.i | + | GB.j | + | GC.k |

<= GA + GB + GC.

Best regards

Nikos Dergiades

> > ABC with centroid G and incenter I, then

___________________________________________________________

> > AG+BG+CG-AI-BI-CI=0 or >

> > 0. Is that true?

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