335. Some difficult geometric/trigonometric inequalities.
by Virgil Nicula, Jan 29, 2012, 3:43 PM
PP1 (USAMO - 2007). Let an acute 
with incircle 
and circumcircle 
. The circles 
and 
are tangent internally
to
in the same 
. 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](//latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of 
.
Proof 1. Prove easily that
; 
. Let 
, 
.
Apply Stewart's theorem in
for 
and 
.
Proof 2. Apply the Pythagoras' theorem in the triangles :
Therefore,
, i.e. 
. Observe that 
, where 
is the 
- exinradius of .
Remark. Prove easily that two more interesting relations :
and 
, 
.
Remark.
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](//latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
.
PP2. Let so that 
. Prove that 
with equality iff 
.
Remark. If
, then our inequality becomes 
with equality iff 
.
Proof. If
, then the inequality 
becomes 
, what is true. We have equality iff 
, i.e. 
. Suppose 
and denote
, i.e. 
, where 
. Inequality becomes 
, what is true.
We have equality iff
, i.e. 
.
Remark. If
, i.e. 
, then becomes 
, with equality iff .
PP3 (Mateescu Constantin). Prove that
with semiperimeter 
there is relation 
,
where
, 
and 
denote the lengths of the Nagel cevians corresponding to the vertices , 
and 
respectively.
Proof 1.
. Denote 
. Observe that 
and 
.
Thus,
. Since 
obtain that
.
In conclusion,
.
Proof 2. I will use the following well known results:
. Consequently: 
![$ \sum\, \left[s^2-\frac {4s(s-b)(s-c)}a\right]+$](//latex.artofproblemsolving.com/6/2/4/624efc771073161071c225fba923d5649817ed5e.png)
![$2\sum\, \sqrt{\left[s^2-\frac {4s(s-c)(s-a)}b\right]\cdot \left[s^2-\frac {4s(s-a)(s-b)}c\right]}=$](//latex.artofproblemsolving.com/5/1/1/511f2801811718bd5c16c9d1d7a9c7d98c615f72.png)
![$3s^2-\sum\, \frac {4s(s-b)(s-c)}a+2s\sum\, \sqrt{\frac {\left[sb-4(s-c)(s-a)\right]\cdot\left[sc-4(s-a)(s-b)\right]}{bc}}\stackrel{(2)}{\ge}$](//latex.artofproblemsolving.com/e/2/a/e2a863f8cd7ceac60e8c9366d48e57c3f4eca0f8.png)
.
Firstly, we will establish the following identity: 
. Let 
be the foot of the Nagel-cevian issued from vertex to the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.
Thus,
and by using the law of cosine in 
one obtains: 
and since 
, the latter
equality rewrites as:
![$s^2-\left(s-c\right)\left[2c+\tfrac {c^2+a^2-b^2}a\right]=$](//latex.artofproblemsolving.com/c/e/7/ce776a54a749e8281dc7b9bf25d701482400919d.png)
, as desired. Of course, the cyclic versions hold as well, so by summing them up we get:
. On the other hand, note that 
![$\sum\, bc\cdot\left[s^2-s\left(b+c\right)+bc\right]=$](//latex.artofproblemsolving.com/b/7/0/b706ee826b78104ed0059487092dca8e64310d98.png)
and therefore our above summation is
finally equal to:
. By the well-known inequality 
, which holds for 
, we now deduce
that:
, so it suffices to show that: 
or 
.
But the latter inequality is weaker than Gerretsen's one i.e.
as can be seen from the fact that the inequality
reduces to the obvious 
.
PP4. Prove that
.
Proof 1 (M.C.). The L.H.S. of the inequality can be written as![$\sum\, \frac {\cos^2A}{1+\cos A}=\sum\, -\left[\frac {1-\cos^2A-1}{1+\cos A}\right]=$](//latex.artofproblemsolving.com/8/1/f/81f16bf823616e1b9561c7f72607001c059e9fbb.png)
. Making use of the following set of identities 
one obtains (after some computations) that
. Therefore, it remains to prove the inequality 
, which
is true since it follows from the remarkable distance![$\boxed{\ H\Gamma^2=4R^2\left[1-\frac{2s^2\left(2R-r\right)}{R\left(4R+r\right)^2}\right]\ }$](//latex.artofproblemsolving.com/6/0/1/60117aef98ff464872ddff93a38f45ed450e466e.png)
, where 
and 
represent the orthocenter and Gergonne's point respectively of 
.
Proof 2. Observe that
. Thus, 
.
![$\sum\frac {\left(b^2+c^2-a^2\right)^2}{bc(b+c-a)}\stackrel{(C.B.S.)}{\ge} \frac {\left[\sum\left(b^2+c^2-a^2\right)\right]^2}{\sum bc(b+c-a)}=\frac {\left(a^2+b^2+c^2\right)^2}{\sum bc(b+c-a)}$](//latex.artofproblemsolving.com/2/7/e/27ebf82aadb8a78b1d1d76b1a14492f676dc1d55.png)
. It remains to prove the inequality 
. Indeed, this relation is equivalently with
![$\sum a^4+2\sum b^2c^2\ge \sum bc\left[(b+c)^2-a^2\right]\iff$](//latex.artofproblemsolving.com/9/7/f/97ffd20b41f22f44557367400a7e68b874a5ceb9.png)
(Schur's inequality).
Proof 3 (own).
Remarks. Here are some equivalent forms of the Schur's inequalities for
and 
, where 
.
, where 
.
. I used 
.
Example. Prove that
==> http://www.artofproblemsolving.com/blog/65909
PP5. Let
be the lengths of the sides for and 
its semiperimeter. Show that
, i.e. 
.
Proof 1 (own). Let
be the incircle for the triangle 
. Denote the second intersections 
of the lines 
respectively with circumcircle 
of the triangle. We remark the following three properties ![$:\ \boxed {A'B=A'C=A'I}\ (1)\ ;\ [AC'BA'CB']=\sum [OBA'C]=$](//latex.artofproblemsolving.com/c/7/3/c73517cc911e07f584ca046d53d88bc5740aa775.png)
![$ \boxed {\ [AC'BA'CB']=pR\ }\ (2)\ ;\ [IBA'C]=\frac 12\cdot a\cdot IA'\cdot \sin \left(B+\frac A2\right)\le $](//latex.artofproblemsolving.com/8/2/c/82c23a533e566df8520e8e3b62929cd846df3b02.png)
![$\boxed {\ a\cdot IA'\ge 2\cdot [IBA'C]\ }\ (3)$](//latex.artofproblemsolving.com/a/f/6/af6cbac7d495c57be41b7ccc025531a6a0e216ce.png)
. In conclusion,
![$\frac{1}{2R}\cdot \sum a\cdot IA'\ge\frac{1}{2R}\cdot \sum 2\cdot [IBA'C]=$](//latex.artofproblemsolving.com/e/0/9/e099ca3afff88a758aab518c74c5c0c879d703c0.png)
![$\frac 1R\cdot \sum [IBA'C]=\frac 1R\cdot [AC'BA'CB']=p\Longrightarrow\boxed {\ \sum a\cdot \sin \frac A2\ge p\ }\ .$](//latex.artofproblemsolving.com/b/c/f/bcf42b8d828310a082e741464203d43addf33f87.png)
Proof 2. I"ll use a simple inequality
and two identities 
. Indeed, 
![$\sum [(a+b+c)-a]\cdot \sin^{2}\frac{A}{2}=$](//latex.artofproblemsolving.com/b/6/4/b64d95320832b623982e4d748cd9f4d80dca6c7f.png)
.
PP6. Let the point
in the interior of . 
intersect the circumcircle of at 
respectively.
Prove that![$[A_{1}BC]\ +\ [B_{1}AC]\ +\ [C_{1}AB]\ \le\ p(R-r)$](//latex.artofproblemsolving.com/3/d/0/3d0b32b57ab71a9e3060cf3cf1619d3a2d53a10f.png)
, where 
, 
, 
are the usual notations for .
Proof. I"ll use the well-known relations
Let
the middlepoints
of the sides , ![$[CA]$](//latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and the middlepoints 
, 
, 
of the arcs 
, 
, 
respectively in the circumcircle. Then 
the arrow of the arc has the length 
a.s.o. ; ![$[BA_1C]\le [BA'C]=$](//latex.artofproblemsolving.com/2/a/7/2a7c5551dbb5cd0694c44374ca2948447534c280.png)
a.s.o. Thus, ![$\sum [BA_1C]\le \frac 12 R\sum a(1-\cos A)=$](//latex.artofproblemsolving.com/e/e/6/ee67b2697baacc28830062cbe07236287b2e51e2.png)
Therefore, ![$\boxed {\ \sum [BA_1C]\le p(R-r)\ }\Longleftrightarrow [AC_1BA_1CB_1]\le pR\ \ (1)\ .$](//latex.artofproblemsolving.com/d/d/f/ddf3dcbf99333da7d0a208ca50b95bf51e7f3434.png)
Remark. Define :
, 
; 
, 
; 
, 
. We observe that 
, 
,
and 
(the trigonometrical form of the Ceva's theorem). Then the inequality 
, i.e. ![$[AC_1BA_1CB_1]\le pR$](//latex.artofproblemsolving.com/2/c/2/2c2c164484cad473a5cb224efa065af389091605.png)
, what is truly. Thus, we obtain an another proof of the proposed
problem. We have the equality in the inequality if and only if 
, i.e. the point is the incentre of the triangle 
PP7. Prove that
.
Proof 1. Denote
, where 
. Thus, 
. In conclusion, 
, what is truly. We have the equality iff 
and 
.
Proof 2. Denote
. Therefore, 
![$0\le (a-b)^2+\left[(a+b)t-2\right]^2$](//latex.artofproblemsolving.com/7/6/a/76a49c39cf02c8740e0e021597f3be8b7fd85418.png)
, what is truly. We have the equality iff 
and 
.
An easy extension. Prove that
.
PP8. Let be a triangle. Denote the inradius of 
. Prove that 
.
Proof. Show easily that
a.s.o. Thus, we have 
. The circumradius of has the length 
.
I"ll use same well-known identity in both triangles
. Observe that 
, what is truly. In conclusion, 
. The Euler's inequality
in
. I"ll use the identities 
. Therefore, 
. Since 
obtain
that
. Obtain equivalently that 
. Apply to the orthic triangle of an acute triangle 
and 
because 
and 
. Remark. Gerretsen's
inequality
.
PP9. we have 
and prove that 
.
Method 0.
, what is truly. We have the equality if and only if 
, i.e. the triangle is -isosceles.
Method 1. Let: the second intersection between circumcircle and bisector 
of 
; the intersection 
. From the relations
and 
because 
results: 
. Therefore, the inequality is equivalently with .
Method 2. Using the identity
results 
![$(p-b)(p-c)(b+c)^2\le a^2 [p(p-a)+(p-b)(p-c)]$](//latex.artofproblemsolving.com/7/f/7/7f79cc27a71d0eefc8ee485ca64fcdd7462f1767.png)
![$(p-b)(p-c)\left[(b+c)^2-a^2\right]\le a^2p(p-a)$](//latex.artofproblemsolving.com/e/8/d/e8dfc4cbde6144270516058c2034594203443fff.png)
.
Remark. Using this lemma, our inequality is a consequence. The equality holds
. Prove easily that .
PP10. Prove that in an acute with the circumcircle 
there is the inequality 
.
Proof. Denote the circumcenter
of and the projections 
, 
of the points , on the sideline 
respectively. Prove easily that 
,
i.e.
and 
. Remain to show that 
. Indeed, 
, i.e. the relation .
Remark. is a application of 
. Remark that 
.
PP11. Prove that
.
Proof. Prove easily
and apply upper trigonometrical inequality 
for 
, 
and 
.
PP12. Prove that in an acute-angled triangle there is the inequality 
.
Let
, 
and 
. Thus 
and prove easily 
, what
is truly. So
.
If denote so that 
, 
and 
, then . Since 
a.s.o. Hence 
with equality iff 
, 
and 
or any cyclic permutation.
PP13. Let with circumcircle 
, its interior and 
. For the same way we get 
, 
, 
, 
. Prove that 
.
Proof. Denote
. Thus, 
![$\frac {2R^2\left[\cos \left(a_1-a_2\right)-\cos A\right]}{bc}\le $](//latex.artofproblemsolving.com/7/5/5/755d2a52b007a478cdd5892083081fc8f1c52907.png)
because 
, what is truly. In conclusion, 
. We have the equality 
- the incenter of .
PP14. Prove that in an acute exists the inequality 
, where 
.
Proof.
, what is truly.
PP15. Prove that in an acute there is the inequality 
.
Proof (arqady).
, where the last inequality is equivalent to ![$\left(\sqrt3\tan\frac{C}{2}-1 \right)^2\left[\left(3-\sqrt3\right)\tan\frac{C}{2}+\left(3\sqrt3-5\right)\right]\ge 0$](//latex.artofproblemsolving.com/b/d/9/bd9bd5adf4f6f0e85af4e95444d35bff158c5aee.png)
, which is obvious.
Remark. Nice proof ! This inequality is equivalent to
with incircle 
and circumcircle 
. The circles 
and 
are tangent internallyto
in the same 
. 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](http://latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of 
.Proof 1. Prove easily that
; 
. Let 
, 
.Apply Stewart's theorem in
for 
and 
.Proof 2. Apply the Pythagoras' theorem in the triangles :
Therefore,
, i.e. 
. Observe that 
, where 
is the 
- exinradius of .Remark. Prove easily that two more interesting relations :
and 
, 
.Remark.
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](http://latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
.PP2. Let
so that 
. Prove that 
with equality iff 
.Remark. If
, then our inequality becomes 
with equality iff 
.Proof. If
, then the inequality 
becomes 
, what is true. We have equality iff 
, i.e. 
. Suppose 
and denote
, i.e. 
, where 
. Inequality becomes 
, what is true.We have equality iff
, i.e. 
.Remark. If
, i.e. 
, then becomes 
, with equality iff .PP3 (Mateescu Constantin). Prove that
with semiperimeter 
there is relation 
,where
, 
and 
denote the lengths of the Nagel cevians corresponding to the vertices , 
and 
respectively.Proof 1.
. Denote 
. Observe that 
and 
.Thus,
. Since 
obtain that
.In conclusion,
.Proof 2. I will use the following well known results:
. Consequently: 
![$ \sum\, \left[s^2-\frac {4s(s-b)(s-c)}a\right]+$](http://latex.artofproblemsolving.com/6/2/4/624efc771073161071c225fba923d5649817ed5e.png)
![$2\sum\, \sqrt{\left[s^2-\frac {4s(s-c)(s-a)}b\right]\cdot \left[s^2-\frac {4s(s-a)(s-b)}c\right]}=$](http://latex.artofproblemsolving.com/5/1/1/511f2801811718bd5c16c9d1d7a9c7d98c615f72.png)
![$3s^2-\sum\, \frac {4s(s-b)(s-c)}a+2s\sum\, \sqrt{\frac {\left[sb-4(s-c)(s-a)\right]\cdot\left[sc-4(s-a)(s-b)\right]}{bc}}\stackrel{(2)}{\ge}$](http://latex.artofproblemsolving.com/e/2/a/e2a863f8cd7ceac60e8c9366d48e57c3f4eca0f8.png)
.
Firstly, we will establish the following identity: 
. Let 
be the foot of the Nagel-cevian issued from vertex to the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.Thus,
and by using the law of cosine in 
one obtains: 
and since 
, the latterequality rewrites as:
![$s^2-\left(s-c\right)\left[2c+\tfrac {c^2+a^2-b^2}a\right]=$](http://latex.artofproblemsolving.com/c/e/7/ce776a54a749e8281dc7b9bf25d701482400919d.png)
, as desired. Of course, the cyclic versions hold as well, so by summing them up we get:
. On the other hand, note that 
![$\sum\, bc\cdot\left[s^2-s\left(b+c\right)+bc\right]=$](http://latex.artofproblemsolving.com/b/7/0/b706ee826b78104ed0059487092dca8e64310d98.png)
and therefore our above summation isfinally equal to:
. By the well-known inequality 
, which holds for 
, we now deducethat:
, so it suffices to show that: 
or 
.But the latter inequality is weaker than Gerretsen's one i.e.
as can be seen from the fact that the inequality
reduces to the obvious 
.PP4. Prove that
. Proof 1 (M.C.). The L.H.S. of the inequality can be written as
![$\sum\, \frac {\cos^2A}{1+\cos A}=\sum\, -\left[\frac {1-\cos^2A-1}{1+\cos A}\right]=$](http://latex.artofproblemsolving.com/8/1/f/81f16bf823616e1b9561c7f72607001c059e9fbb.png)
. Making use of the following set of identities 
one obtains (after some computations) that
. Therefore, it remains to prove the inequality 
, whichis true since it follows from the remarkable distance
![$\boxed{\ H\Gamma^2=4R^2\left[1-\frac{2s^2\left(2R-r\right)}{R\left(4R+r\right)^2}\right]\ }$](http://latex.artofproblemsolving.com/6/0/1/60117aef98ff464872ddff93a38f45ed450e466e.png)
, where 
and 
represent the orthocenter and Gergonne's point respectively of 
.Proof 2. Observe that
. Thus, 
.![$\sum\frac {\left(b^2+c^2-a^2\right)^2}{bc(b+c-a)}\stackrel{(C.B.S.)}{\ge} \frac {\left[\sum\left(b^2+c^2-a^2\right)\right]^2}{\sum bc(b+c-a)}=\frac {\left(a^2+b^2+c^2\right)^2}{\sum bc(b+c-a)}$](http://latex.artofproblemsolving.com/2/7/e/27ebf82aadb8a78b1d1d76b1a14492f676dc1d55.png)
. It remains to prove the inequality 
. Indeed, this relation is equivalently with
![$\sum a^4+2\sum b^2c^2\ge \sum bc\left[(b+c)^2-a^2\right]\iff$](http://latex.artofproblemsolving.com/9/7/f/97ffd20b41f22f44557367400a7e68b874a5ceb9.png)
(Schur's inequality).Proof 3 (own).
Remarks. Here are some equivalent forms of the Schur's inequalities for
and 
, where 
.
, where 
.
. I used 
.Example. Prove that
==> http://www.artofproblemsolving.com/blog/65909PP5. Let
be the lengths of the sides for and 
its semiperimeter. Show that
, i.e. 
.Proof 1 (own). Let
be the incircle for the triangle 
. Denote the second intersections 
of the lines 
respectively with circumcircle 
of the triangle
. We remark the following three properties ![$:\ \boxed {A'B=A'C=A'I}\ (1)\ ;\ [AC'BA'CB']=\sum [OBA'C]=$](http://latex.artofproblemsolving.com/c/7/3/c73517cc911e07f584ca046d53d88bc5740aa775.png)
![$ \boxed {\ [AC'BA'CB']=pR\ }\ (2)\ ;\ [IBA'C]=\frac 12\cdot a\cdot IA'\cdot \sin \left(B+\frac A2\right)\le $](http://latex.artofproblemsolving.com/8/2/c/82c23a533e566df8520e8e3b62929cd846df3b02.png)
![$\boxed {\ a\cdot IA'\ge 2\cdot [IBA'C]\ }\ (3)$](http://latex.artofproblemsolving.com/a/f/6/af6cbac7d495c57be41b7ccc025531a6a0e216ce.png)
. In conclusion,
![$\frac{1}{2R}\cdot \sum a\cdot IA'\ge\frac{1}{2R}\cdot \sum 2\cdot [IBA'C]=$](http://latex.artofproblemsolving.com/e/0/9/e099ca3afff88a758aab518c74c5c0c879d703c0.png)
![$\frac 1R\cdot \sum [IBA'C]=\frac 1R\cdot [AC'BA'CB']=p\Longrightarrow\boxed {\ \sum a\cdot \sin \frac A2\ge p\ }\ .$](http://latex.artofproblemsolving.com/b/c/f/bcf42b8d828310a082e741464203d43addf33f87.png)
Proof 2. I"ll use a simple inequality
and two identities 
. Indeed, 
![$\sum [(a+b+c)-a]\cdot \sin^{2}\frac{A}{2}=$](http://latex.artofproblemsolving.com/b/6/4/b64d95320832b623982e4d748cd9f4d80dca6c7f.png)
.PP6. Let the point
in the interior of . 
intersect the circumcircle of at 
respectively.Prove that
![$[A_{1}BC]\ +\ [B_{1}AC]\ +\ [C_{1}AB]\ \le\ p(R-r)$](http://latex.artofproblemsolving.com/3/d/0/3d0b32b57ab71a9e3060cf3cf1619d3a2d53a10f.png)
, where 
, 
, 
are the usual notations for .Proof. I"ll use the well-known relations
Letthe middlepoints
of the sides , ![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and the middlepoints 
, 
, 
of the arcs 
, 
, 
respectively in the circumcircle. Then 
the arrow of the arc
has the length 
a.s.o. ; ![$[BA_1C]\le [BA'C]=$](http://latex.artofproblemsolving.com/2/a/7/2a7c5551dbb5cd0694c44374ca2948447534c280.png)
a.s.o. Thus, ![$\sum [BA_1C]\le \frac 12 R\sum a(1-\cos A)=$](http://latex.artofproblemsolving.com/e/e/6/ee67b2697baacc28830062cbe07236287b2e51e2.png)
Therefore, ![$\boxed {\ \sum [BA_1C]\le p(R-r)\ }\Longleftrightarrow [AC_1BA_1CB_1]\le pR\ \ (1)\ .$](http://latex.artofproblemsolving.com/d/d/f/ddf3dcbf99333da7d0a208ca50b95bf51e7f3434.png)
Remark. Define :
, 
; 
, 
; 
, 
. We observe that 
, 
,
and 
(the trigonometrical form of the Ceva's theorem). Then the inequality 
, i.e. ![$[AC_1BA_1CB_1]\le pR$](http://latex.artofproblemsolving.com/2/c/2/2c2c164484cad473a5cb224efa065af389091605.png)
, what is truly. Thus, we obtain an another proof of the proposedproblem. We have the equality in the inequality
if and only if 
, i.e. the point is the incentre of the triangle 
PP7. Prove that
.Proof 1. Denote
, where 
. Thus, 
. In conclusion, 
, what is truly. We have the equality iff 
and 
.Proof 2. Denote
. Therefore, 
![$0\le (a-b)^2+\left[(a+b)t-2\right]^2$](http://latex.artofproblemsolving.com/7/6/a/76a49c39cf02c8740e0e021597f3be8b7fd85418.png)
, what is truly. We have the equality iff 
and 
.An easy extension. Prove that
.PP8. Let
be a triangle. Denote the inradius of 
. Prove that 
. Proof. Show easily that
a.s.o. Thus, we have 
. The circumradius of has the length 
.I"ll use same well-known identity in both triangles
. Observe that 
, what is truly. In conclusion, 
. The Euler's inequalityin
. I"ll use the identities 
. Therefore, 
. Since 
obtainthat
. Obtain equivalently that 
. Apply to the orthic triangle of an acute triangle 
and 
because 
and 
. Remark. Gerretsen'sinequality
.PP9.
we have 
and prove that 
.Method 0.
, what is truly. We have the equality if and only if 
, i.e. the triangle is -isosceles.Method 1. Let: the second intersection
between circumcircle and bisector 
of 
; the intersection 
. From the relations
and 
because 
results: 
. Therefore, the inequality is equivalently with .Method 2. Using the identity
results 
![$(p-b)(p-c)(b+c)^2\le a^2 [p(p-a)+(p-b)(p-c)]$](http://latex.artofproblemsolving.com/7/f/7/7f79cc27a71d0eefc8ee485ca64fcdd7462f1767.png)
![$(p-b)(p-c)\left[(b+c)^2-a^2\right]\le a^2p(p-a)$](http://latex.artofproblemsolving.com/e/8/d/e8dfc4cbde6144270516058c2034594203443fff.png)
.Remark. Using this lemma, our inequality is a consequence. The equality holds
. Prove easily that .PP10. Prove that in an acute
with the circumcircle 
there is the inequality 
. Proof. Denote the circumcenter
of and the projections 
, 
of the points , on the sideline 
respectively. Prove easily that 
,i.e.
and 
. Remain to show that 
. Indeed, 
, i.e. the relation .Remark.
is a application of 
. Remark that 
.PP11. Prove that
.Proof. Prove easily
and apply upper trigonometrical inequality 
for 
, 
and 
.PP12. Prove that in an acute-angled triangle
there is the inequality 
.Let
, 
and 
. Thus 
and prove easily 
, whatis truly. So
.If denote
so that 
, 
and 
, then . Since 
a.s.o. Hence 
with equality iff 
, 
and 
or any cyclic permutation.PP13. Let
with circumcircle 
, its interior and 
. For the same way we get 
, 
, 
, 
. Prove that 
.Proof. Denote
. Thus, 
![$\frac {2R^2\left[\cos \left(a_1-a_2\right)-\cos A\right]}{bc}\le $](http://latex.artofproblemsolving.com/7/5/5/755d2a52b007a478cdd5892083081fc8f1c52907.png)
because 
, what is truly. In conclusion, 
. We have the equality 
- the incenter of .PP14. Prove that in an acute
exists the inequality 
, where 
.Proof.
, what is truly.PP15. Prove that in an acute
there is the inequality 
.Proof (arqady).
, where the last inequality is equivalent to ![$\left(\sqrt3\tan\frac{C}{2}-1 \right)^2\left[\left(3-\sqrt3\right)\tan\frac{C}{2}+\left(3\sqrt3-5\right)\right]\ge 0$](http://latex.artofproblemsolving.com/b/d/9/bd9bd5adf4f6f0e85af4e95444d35bff158c5aee.png)
, which is obvious.Remark. Nice proof ! This inequality is equivalent to
This post has been edited 184 times. Last edited by Virgil Nicula, Nov 19, 2015, 11:54 AM
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