224. Some nice and easy inequalities in a triangle.
by Virgil Nicula, Feb 18, 2011, 4:30 PM
Proposed problem 1. Let 
, and 
be the side lengths of a triangle 
. Let 
, 
and 
be
the inradius, the exradii and the semiperimeter of triangle . Prove that 
.
Proof. Here you can find many identities, for example
.
I am using them in solving inequalities. Therefore for this well-known inequality appear at least three methods :
.
because 
.
![$\blacktriangleright\ \sum a^2r_a=\frac 1r\cdot\sum\left[a\cdot a(s-b)(s-c)\right]\ \stackrel{\mathrm{Chebychev}\uparrow\uparrow}{\ge}\ \frac {1}{3r}\cdot$](//latex.artofproblemsolving.com/b/1/5/b159c6e95bf5cd6975c83a5feb410b9befec69b3.png)
because 
.
Proposed problem 2. Let be a triangle with incircle 
and circumcircle 
. Consider 
, 
,
for which denote 
, 
, 
. Prove that
with equality iff 
.
Proof. Apply the Stewart's relation for cevian
and 
, with equality iff 
, in other words, equality is attained when is the 
- bisector .
Since
. Thus, the last inequality becomes 
,
i.e.
. Now, adding up the similar inequalities yields the desired result . Equality holds iff .
Lemma. Let
with orthocenter 
and circumcircle 
. Then there is the inequality 
.
Euler's relation. Suppose w.l.o.g.
. Then 
,
where denote the distance
of the point 
to the line 
and the length 
of the 
-inradius. Remark : 
.
Proposed problem 3. In any triangle there is the inequality 
.
Proof. From the well-known property (in any triangle)
a.s.o., from the Euler's relation and using the above lemma obtain
because
and multiply with 
and multiply
with
and multiply with 
. Hence the inequality 
is truly.
Proposed problem 4.![\[\begin{array}{c}
\\
\boxed{\ \sum \cos^3\frac{A}{2} \cdot \cos \frac{B-C}{2}\le \frac{9\sqrt{3 }}{8}\ }\\\\
\sum 2\cos^2\frac A2\cdot 2\sin\frac {B+C}{2}\cos\frac {B-C}{2}\le\frac {9\sqrt 3}{2}\\\\
\sum (1+\cos A)(\sin B+\sin C)\le\frac{9\sqrt{3 }}{2}\\\\
2\sum \sin A+\sum (\sin B\cos C+\sin C\cos B)\le\frac{9\sqrt{3 }}{2}\\\\
2\sum \sin A+\sum \sin (B+C)\le\frac{9\sqrt{3 }}{2}\\\\
3\sum \sin A\le\frac{9\sqrt{3 }}{2}\\\\
\sum \sin A\le \frac {3\sqrt 3}{2}\\\\
\sum\frac {a}{2R}\le\frac {3\sqrt 3}{2}\\\\
\frac sR\le \frac {3\sqrt 3}{2}\\\\
s\le \frac {3R\sqrt 3}{2}\\\\
\mathrm{O.K.}\\\
\end{array}\]](//latex.artofproblemsolving.com/e/7/6/e76a947ae6f9b12d127b42c678fa2407a156b395.png)
Proposed problem 5. Prove that
, where 
.
Proof.
a.s.o. 
. Therefore, 
. Remark that 
because 
. In conclusion, 
. Nice !
Proposed problem 6. Prove that
, where 
is the length of the -median a.s.o.
Proof. Denote the centroid
and the points 
, 
, 
. Apply the Ptolemy's inequality
in the quadrilaterals
.
, and 
be the side lengths of a triangle 
. Let 
, 
and 
bethe inradius, the exradii and the semiperimeter of triangle
. Prove that 
.Proof. Here you can find many identities, for example
.I am using them in solving inequalities. Therefore for this well-known inequality appear at least three methods :
.
because 
.![$\blacktriangleright\ \sum a^2r_a=\frac 1r\cdot\sum\left[a\cdot a(s-b)(s-c)\right]\ \stackrel{\mathrm{Chebychev}\uparrow\uparrow}{\ge}\ \frac {1}{3r}\cdot$](http://latex.artofproblemsolving.com/b/1/5/b159c6e95bf5cd6975c83a5feb410b9befec69b3.png)
because 
.Proposed problem 2. Let
be a triangle with incircle 
and circumcircle 
. Consider 
, 
,
for which denote 
, 
, 
. Prove that
with equality iff 
.Proof. Apply the Stewart's relation for cevian
and 
, with equality iff 
, in other words, equality is attained when is the 
- bisector .Since
. Thus, the last inequality becomes 
,i.e.
. Now, adding up the similar inequalities yields the desired result . Equality holds iff .Lemma. Let
with orthocenter 
and circumcircle 
. Then there is the inequality 
.
, 
and power of 
w.r.t. 
is 
.
.
obtain 
.Euler's relation. Suppose w.l.o.g.
. Then 
,where denote the distance
of the point 
to the line 
and the length 
of the 
-inradius. Remark : 
.
, 
, 
of the sides ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
, ![$ [CA]$](http://latex.artofproblemsolving.com/b/8/3/b8309224981a62a2c0df087562b92830bd7ad7b0.png)
, ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
respectively and 
, 
,
. I"ll apply the Ptolemeu's relation in the quadrilaterals 
, 
, 
.
. ![$ \left\|\begin{array}{cccc} OMBP\ : & cx + az & = & bR \\
\ ONCM\ : & ay + bx & = & cR \\
\ OPAN\ : & bz + cy & = & aR \\
\ [ABC] = sr & ax + by + cz & = & 2sr\end{array}\right\|\bigoplus\ \implies\ 2s(x + y + z) = 2s(R + r)$](http://latex.artofproblemsolving.com/d/d/9/dd9877466d09ad5c850e94294f3de80d2914c6ab.png)
.
. ![$ \left\|\begin{array}{cccc} OMBP\ : & cx + bR & = & az \\
\ OMNC\ : & bx + cR & = & ay \\
\ OPAN\ : & cy + bz & = & aR \\
\ [ABC] = (s - a)r_a & by + cz - ax & = & 2(s - a)r_a\end{array}\right\|\bigoplus\ \implies$](http://latex.artofproblemsolving.com/2/e/f/2ef4d7a0dc0d5d8147cc9517d7e7fa56d6bb7f02.png)
.
, then 
, 
, 
, 
, 
, 
, 
, 
and 
.Proposed problem 3. In any triangle
there is the inequality 
.Proof. From the well-known property (in any triangle)
a.s.o., from the Euler's relation and using the above lemma obtain
because
and multiply with 
and multiplywith
and multiply with 
. Hence the inequality 
is truly.Proposed problem 4.
![\[\begin{array}{c}
\\
\boxed{\ \sum \cos^3\frac{A}{2} \cdot \cos \frac{B-C}{2}\le \frac{9\sqrt{3 }}{8}\ }\\\\
\sum 2\cos^2\frac A2\cdot 2\sin\frac {B+C}{2}\cos\frac {B-C}{2}\le\frac {9\sqrt 3}{2}\\\\
\sum (1+\cos A)(\sin B+\sin C)\le\frac{9\sqrt{3 }}{2}\\\\
2\sum \sin A+\sum (\sin B\cos C+\sin C\cos B)\le\frac{9\sqrt{3 }}{2}\\\\
2\sum \sin A+\sum \sin (B+C)\le\frac{9\sqrt{3 }}{2}\\\\
3\sum \sin A\le\frac{9\sqrt{3 }}{2}\\\\
\sum \sin A\le \frac {3\sqrt 3}{2}\\\\
\sum\frac {a}{2R}\le\frac {3\sqrt 3}{2}\\\\
\frac sR\le \frac {3\sqrt 3}{2}\\\\
s\le \frac {3R\sqrt 3}{2}\\\\
\mathrm{O.K.}\\\
\end{array}\]](http://latex.artofproblemsolving.com/e/7/6/e76a947ae6f9b12d127b42c678fa2407a156b395.png)
Proposed problem 5. Prove that
, where 
. Proof.
a.s.o. 
. Therefore, 
. Remark that 
because 
. In conclusion, 
. Nice ! Proposed problem 6. Prove that
, where 
is the length of the -median a.s.o. Proof. Denote the centroid
and the points 
, 
, 
. Apply the Ptolemy's inequalityin the quadrilaterals
.This post has been edited 29 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:06 PM
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