Original proposal of Mathematical Re‡ections problem O25 / Darij Grinberg
The following problem submission made it into the periodical "Mathematical Re-
‡ections" as Problem O25 (in a shortened form). Below is my original solution of this problem. A much simpler solution was published in "Mathematical Re‡ections" issue 6/2006.
Problem. For any triangle ABC; prove that cosA
2 cotA
2 + cosB 2 cotB
2 + cosC 2 cotC
2
p3
2 cotA
2 + cotB
2 + cotC 2 9
2 2 cos2A
2 + cos2 B
2 + cos2C 2 : Solution. The only interesting part of the inequality is
cosA 2 cotA
2 + cosB 2 cotB
2 + cosC 2 cot C
2
p3
2 cotA
2 + cotB
2 + cotC
2 ; (1) because the other two parts of the inequality are pretty easy:
Since the angles of a triangle sum up to180 ;we have A+B+C = 180 ;since the functionf(x) = cotxis convex on the interval ]0 ; 90 [(the interval where the angles
A 2; B
2; C
2 lie), the Jensen inequality yields
cotA
2+cotB
2+cotC
2 3 cot A
2 + B 2 +C
2
3 = 3 cotA+B+C
6 = 3 cot180
6 = 3 cot 30 = 3p 3;
so that p
3
2 cotA
2 + cotB
2 + cotC 2
9 2; and sincecos' = 2 cos2 '
2 1for every angle'; the famous triangle inequalitycosA+ cosB + cosC 3
2 rewrites as 2 cos2 A
2 1 + 2 cos2 B
2 1 + 2 cos2 C
2 1 3
2; so that
2 cos2A
2 + cos2B
2 + cos2 C
2 3 3
2; and thus 2 cos2 A
2 + cos2 B
2 + cos2 C
2 3 + 3 2 = 9
2:
So it only remains to prove the inequality (1). Lets= a+b+c
2 be the semiperime- ter of triangle ABC: Then, it is known that the reals x= s a; y = s b; z =s c are positive. Also, x+y+z = s (because x+y+z = (s a) + (s b) + (s c) = 3s (a+b+c) = 3s 2s = s) and y+z = a (because y+z = (x+y+z) x =
1
s (s a) =a) and similarlyz+x=bandx+y=c:Hence, the well-known half-angle formulascosA
2 =
rs(s a)
bc andcotA 2 =
s s(s a)
(s b) (s c) (the latter is better known in the equivalent form tanA
2 =
s(s b) (s c)
s(s a) ) rewrite as
cosA 2 =
s
(x+y+z)x
(z+x) (x+y) and cotA 2 =
s
(x+y+z)x yz : Now, using the signP
for cyclic sums, the inequality (1) becomes XcosA
2 cotA 2
p3 2
XcotA 2; but
XcosA 2 cotA
2 =Xs
(x+y+z)x (z+x) (x+y)
s
(x+y+z)x
yz =X (x+y+z)x p(z+x) (x+y)yz
and
XcotA
2 =Xs
(x+y+z)x
yz =
rx+y+z xyz
Xx=
rx+y+z
xyz (x+y+z); so this inequality becomes
X (x+y+z)x p(z+x) (x+y)yz
p3 2
rx+y+z
xyz (x+y+z):
Upon multiplication by
pxyz
x+y+z;this rewrites as X xp
p x
(z+x) (x+y)
p3 2
px+y+z; or, equivalently, X x2
px(z+x) (x+y)
p3 2
px+y+z:
Now, by the Cauchy-Schwarz inequality in Engel form, X x2
px(z+x) (x+y)
(x+y+z)2 P px(z+x) (x+y);
so it remains to prove that
(x+y+z)2 P px(z+x) (x+y)
p3 2
px+y+z:
2
This simpli…es to q
(x+y+z)3
p3 2
X px(z+x) (x+y):
Squaring this yields
(x+y+z)3 3 4
X px(z+x) (x+y)
2
; i. e.
4 (x+y+z)3 3 X p
x(z+x) (x+y)
2
:
This rewrites as
4 (x+y+z)3 3 X
x(z+x) (x+y) + 2X p
y(x+y) (y+z) p
z(y+z) (z+x) ; i. e.
4 (x+y+z)3 3 X
x(z+x) (x+y) + 2X
(y+z)p
yz(z+x) (x+y) ; i. e.
4 (x+y+z)3 3X
x(z+x) (x+y) + 6X
(y+z)p
yz(z+x) (x+y); i. e.
4 (x+y+z)3 3X
x(z+x) (x+y) 6X
(y+z)p
yz(z+x) (x+y):
Now,
4 (x+y+z)3 3X
x(z+x) (x+y)
= 4 (x+y+z)3 3X
((x+y+z) (z+x) (x+y) (y+z) (z+x) (x+y))
= 4 (x+y+z)3 3 (x+y+z)X
(z+x) (x+y) + 9 (y+z) (z+x) (x+y)
= (x+y+z) 4 (x+y+z)2 3X
(z+x) (x+y) + 9 (y+z) (z+x) (x+y)
= (x+y+z) 0
@x2+y2+z2 yz zx xy
| {z }
0;as you know
1
A+ 9 (y+z) (z+x) (x+y)
9 (y+z) (z+x) (x+y) = 3X
(y+z) (z+x) (x+y)
= 3 X
y(z+x) (x+y) +X
z(z+x) (x+y)
= 3 X
z(x+y) (y+z) +X
y(y+z) (z+x)
= 3X
(z(x+y) (y+z) +y(y+z) (z+x)) = 3X
(y+z) (y(z+x) +z(x+y)); so that, in order to prove the above inequality, it will be enough to show that
3X
(y+z) (y(z+x) +z(x+y)) 6X
(y+z)p
yz(z+x) (x+y); or, equivalently, X(y+z) (y(z+x) +z(x+y)) 2X
(y+z)p
yz(z+x) (x+y):
But this is obvious, since AM-GM yieldsy(z+x)+z(x+y) 2p
y(z+x) z(x+y) = 2p
yz(z+x) (x+y):Thus, the proof of inequality (1) is complete, and the problem is solved.
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