THE HADWIGER FINSLER INEQUALITY REVERSE IN AN ACUTE TRIANGLE sorin radulescu, marian dinca, marius dragan, eduard puschin

THE HADWIGER-FINSLER INEQUALITY REVERSE IN ANACUTE TRIANGLE Prof.dr.. Marian Dincă,Prof.Marius Drăgan,Prof.. Eduard Puschin In the paper1 the authors proposed the following inequality:

THE HADWIGER-FINSLER INEQUALITY REVERSE IN AN

ACUTE TRIANGLE

Prof.dr Sorin Rădulescu,Prof Marian Dincă,Prof.Marius Drăgan,Prof.

Eduard Puschin

In the paper1 the authors proposed the following inequality:

In any acute triangle ABC are true the following inequality:

a2+b2+c2 ≤ 4 3 F + 2− 3

Abstract: In this paper for begining we shall prove the inequality 1 for the isosceles acute triangle

and then for any acute triangle.

Theorem 1.In any isoscel acute triangle ABC with the sides of lengths a, b, c

and area F and b = c are true the inequality 1

Proof :The inequality 1 may be written as :

a2 +b2 +c2

−4 3 F

a−b2+b−c2+c−a2 ≤ 2− 3

Using the notation b = c = x, because F = 12x2sinA and a = 2xsin A2, the right side ratio of inequality 2

may be written as

a2+b2+c2

−4 3 F

a−b2+b−c2+c−a2 = 4x2 sin2 A

2+2x2

−2 3 x2sinA

2 2xsin A

2−x

2 = 4sin2 A

2+2−2 3 sinA

2 2sinA

2 −1

2 = 21−cosA+2−2 3 sinA

2 2sinA

2 −1

2

; 3

= 4−2 cosA+ 3 sinA

2 2sinA

2−1

2 =

4−4 cosA⋅1+ 3

2 ⋅sinA

2 2sinA

2−1

2 = 4 1−cos A−π

3

8 sinA

2− 1 2

2 =

8sin 2 A

2 −π6

8 sinA

2−sin

π

6

2 =

32sin 2 A

4 − 12π cos 2 A

4 − 12π

32sin2 A

4−

π

12 cos 2 A

4+

π

12

=

cos 2 A

4 − 12π

cos 2 A

4+12π

= 1+cos A

2 −π6

1+cos A

2+π6

= 1+cos A

2 −π6

1+cos A

2+π6

− 1 +1 = cos

A

2 −π6 −cos A

2+π6

1+cos A

2+π6

+1

We shall consider the function f : 0, π2 → R, fA = 2sin

A

2 sin π

6

1+cosA

2+π6  +1

As A ≤ π2 , we have A2 + π

6 ≤ π

4 + π

6 = 3π

12 + 2π

12 = 5π

12 <

π

2, or in on equivalent form cos A2 + π

4 + π

6 also sinA2 ≤sinπ

4

it follow that fA ≤ fπ

2 and we have the inequality of the statement

Theorem 2 For any triangle ABC of semiperimeter s with incircle CI, r and circumcicle

CO, R there exist two isosceles triangles

A1B1C1and A2B2C2of semiperimeter s1and s2with incircle CI, r and circumcicle CO, R

Proof The straight line OI cat the circumcircle in it A1and A2.The tangents to the incircle from a point A1, cat the circumcicle in

B1and C1.The straight line B1C1is tangent to the incircle ,acording to Poncelet’s closure theorem.

The triangle A1B1C1is isosceles Similarly the tangents to the incircle from a point A2cat the circumcircle in B2and C2

The straght line B2C2is tangent to the incircle ,acording to Poncelet’s closure theorem it

follows that A2B2C2is isoscel.

As a1 = B1C1 = 2rR+r+d

R+d2

−r2 , b1 = c1 = A1C1 = A1B1 = R+r+dR+d

R+d2

−r2 , where

d =∣ OI ∣= R2− 2Rr and

a2 = B2C2 = 2rR+r−d

R−d2

−r2 , b2 = c2 = A2C2 = A2B2 = R+r−dR−d

R−d2

−r2 and as

s1 = 1a1+b1+c1 = 2R2+10Rr − r2+2R − 2r R2− 2Rr

s2 = 2a2+b2+c2 = 2R2+10Rr − r2− 2R − 2r R2− 2Rr

acording with the W.J.Blundon inequality it follows that the inequality 4 is true

Proof of the inequality (1) If the point O exterior of the incircle C(I,r),rezult∣ OI∣= d ≥ r

, and the tangent to the incircle from a point O,cat

the circumcircle in B3and C3and the tangent to the incircle C(I,r) from a points B3and

C3cat the circumcircle C(O,R) in A3,

acording the Poncelet’s closure theorem.The triangle A3B3C3it is right angled

∢B3 A3C3 = π2,

Let s the semiperimeter of the acute angled triangle ABC and s3the semiperimeter of the triangle A3B3C3

Let the triangle A3B3C3tangent to the incircle C(I,r) in the points ; D.E.F ;where

D ∈ B3C3; E ∈ C3A3and F ∈ B3A3

we have A3F = A3E = x = rctgπ

4 = r, B3F = B3D = y = rctg B3

2 ,

C3D = C3E = z = rctg C3

2 ,

As B3C3 = B3D + C3D = y + z = 2R ,we shall obtain :s3=x + y + z = 2R + r ,it rezults that the inequality 5 is equivalent to :

R∑

ciclic

sinA ≥ r + 2R ,or ∑

ciclic

sinA ≥ R r +2 = ∑

ciclic

cosA + 1, or∑

ciclic

sinA −cosA ≥ 1 ,6 lets

A = π−α2 , B = π−β2 , C = π−γ2 , as A, B, C ∈ 0, π2

we have that α, β, γ ∈ 0, π ,and α + β + γ = π The inequality 6 is equivalented

to∑

ciclic

cosα

2 −sinα2 ≥ 1,or ∑

ciclic

cosα

2 −tgπ4sinα2 ≥ 1, or

∑

ciclic

cosα2 + π

4 ≥cosπ

4 = u, β2 + π

4 = v and γ2 + π

4 = w and using well-know identity:

cosu +cosv +cosw +cosu + v + w = 4∏

ciclic

cos u+v

2  it follows that:

∑

ciclic

cosα

2 +

π

4 +cos∑

ciclic

α

2 + π

4 = 4 ∏

ciclic

cosα+β4 + π

4 = 4 ∏

ciclic

cosπ−γ4 + π

4

or ∑

ciclic

cosα2 + π

4 −cosπ

ciclic

cosπ2 − γ

4 = 4 ∏

ciclic

sinγ4 ≥ 0 we shall obtain the

inequality 7

Using well-known identity: ab + bc + ca = s2+r2+4Rr and

a2+b2+c2 = 2s2− 2r2− 8Rr and F = sr, let 2− 3

3−2 2

= λ

The inequality 1 is equivalented to

4 3 sr + 2λ − 12s2− 2r2− 8Rr − 2λs2+r2+4Rr ≥ 0

or λ − 1s2+2 3 sr + 1 − 3λr2+4 − 12λRr = Es, R, r ≥ 0 8

As λ > 1, we shall obtain : Es, R, r ≥ Es3, R, r 9

In order to prove inequality 8 it will be sufficient to prove inequality Es3, R, r ≥ 0 ,what reprezent the inequality 1 for right angled triangle A3B3C3

We have F = bc2 , and a2 = b2+c2 .We shall obtain the following inequality:

2 3 bc + 2λ − 12b2+c2 − 2λ b2+c2b + c + bc ≥ 0, let b

b+c = x and b+c c = y obtain

2 3 xy + 2λ − 121 − 2xy − 2λ 1 − 2xy + xy ≥ 0 or

xy 3 + 2 − 5λ + 2λ − 1 ≥ λ 1 − 2xy ,let xy = t

As x + y = 1, we shall obtain : t ∈ 0; 14 ,let 1 − 2xy = θ ,θ ∈  1

2 , 1, it will rezult that :xy = 1−θ 2

2 ,we shall obtain:

1−θ 2

2  3 + 2 − 5λ + 2λ − 1 − λ ⋅ θ ≥ 0 or 5λ − 3 − 2θ2− 2λθ − λ + 3 ≥ 0 ,

the equation 5λ − 3 − 2θ2− 2λθ − λ + 3 = 0 has a root θ1 = 1

2 and θ1θ2 = −λ+ 3

5λ− 3 −2

it follows that θ2 = 2 −λ+ 3 

5λ− 3 −2

and θ2 <

2 and 5λ − 3 − 2θ2− 2λθ − λ + 3 = 5λ − 3 − 2θ − θ1θ − θ2 ≥ 0

For ∣ OI ∣= d ≤ r, it shall rezult that any triangle is acute, let ∢B1A1C1 = α and

∢B2 A2C2 = β

sinα2 = r

R+d <

1 2

=sinπ4 , it follows that α2 < π

4 and α < π2 .As sinβ2 = r

2 or

r 2 ≤ R − d , or d ≤ R − r 2 , or

d2 ≤ R − r 2 2or R2− 2Rr ≤ R2− 2 2 Rr + 2r2or  2 − 1R ≤ r or R ≤  2 + 1r ,this inequality is true,because

d ≤ r imply R2− 2Rr ≤ r2, or R − r2 ≤ 2r2 who imply R ≤  2 + 1r we shall obtain:

sinβ2 ≤ 1

2

=sinπ

4 , it follows that β ≤ π

2

Because Es, R, r ≥ Es2, R, r and Es2, R, r ≥ 0 because is an inequality for acute triangle and isosceles triangle

We have inequality of the statement

References

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Finsler-Hadwiger

reverse inequality,Gazeta Matematica seria A,nr 1-2(2010)p.49-53

2 P.G.Popescu,I.V.Maftei,J.L.Diaz-Barrero,M.Dincă:Inegalitati Matematice;Modele

Inovatoare,

Editura Didactica şi Pedagogica,2007

3 Marian Dincă,J.L.Diaz-Barrero: A new proof of an Inequality of Oppenheim, vixra

org:1008.0013

4 Marian Dincă,Mihaly Bencze:New generalisation Finsler-Hadwiger inequality:Octogon

Mathematical Magazine,2002