289. Nice identities in an algebra/ geometry/trigonometry.
by Virgil Nicula, Jun 23, 2011, 1:23 AM
PP1. Suppose that 
such that 
. Prove that 
.
Proof. Are well-known the Viete's relations
. Denote 
, where 
. Observe that 
. Since 
obtain that
. In conclusion, 
.
Particular case. For any
exists the inequality ![$(x+1)^2+x^2+1\ \ge\ 3\cdot\sqrt [3]{2x^2(x+1)^2}$](//latex.artofproblemsolving.com/5/a/9/5a9986f31556e2f609507e3ba64e846b3f270e63.png)
.
PP2. Prove that in a
triangle we have the relation 
.
Proof 1.
.
Remark. I used the well-known identity
which is equivalently with the identity 
.
Indeed, from the relation
obtain that 
. I supposed well-known that 
.
Proof 2.
.
Remark. I used the well-known identity
. Indeed, 
.
Proof 3.
.
Proof 4.
.
Remark.
. Define 
. Then 
.
Indeed, the first relation becomes
and the second relation becomes
.
PP3. Ascertain the monic polynomial equation which has the roots
.
Proof. Using the remarkable sum
obtain that 
. Otherwise. 
.
![$-2\left[\left(\cos\frac {3\pi}{7}+\cos\frac {\pi}{7}\right)-\cos\frac {2\pi}{7}\right]\stackrel{(*)}{=}$](//latex.artofproblemsolving.com/a/f/8/af8ab4edca0b57b1416fc1c6efb2666f28fec0c7.png)
![$-2\left[\left(\frac 12-\cos\frac {5\pi}{7}\right)+\cos\frac {2\pi}{7}\right]=$](//latex.artofproblemsolving.com/d/8/4/d841fa8275f02ad8fb916d77db25798a4d742958.png)
. 
.
In conclusion, are the roots of the equation 
, i.e. 
.
PP4. Let
be a convex quadrilater with 
and 
. Prove that 
.
Proof 1 ("slicing"). . I"ll use the well-known lemma "Let
be a convex quadrilateral so that 
. Then 
" (you can prove easily).
Denote
so that 
. Since 
and 
obtain from upper lemma
that
, i.e. 
. Since 
and 
obtain again from
upper lemma that
, i.e. 
. In conclusion, and 
, i.e. 
.
Proof 2 ("slicing"). Let's draw equilateral triangle
,where the poiny 
lies on the same semi-plane with the point 
w.r.t. 
. Hence 
. Hence 
the streight line 
is the symmetric line of side 
. Then 
. Let
. Hence 
. The symmetric
line of the side will pass throught the points 
and . 
and hence 
. Let's draw
a line dividing
into two equal parts and let's this line crosses the straight line 
in point 
. Hence 
.Then
will be an isosceles triangle 
. 
and the triangle 
will be an isosceles triangle.
, but at the same time and 
. Then 
and
. Let 
. Then in 
.
Proof 3 (metric). quadrilater is orthodiagonal if and only if 
. I"ll use the Sinus' theorem :
.
Replacing in
the last 4 relations we only have to prove that 
.
.
I used the well-known identities
.
Proof 4 (trigonometric - brute force). Denote
. Thus, 
. Apply an well-known identity
. Therefpre,
![$\cos 40^{\circ}[\cos 10^{\circ}-\cos (10^{\circ}+2x)]=\cos 10^{\circ}[\sin 50^{\circ}+\sin (30^{\circ}+2x)]\iff$](//latex.artofproblemsolving.com/3/e/0/3e086bacb8d55ccb54649f01bdaf21b97463f303.png)
Remark. From the step
can continue thus :
![$ [\cos (50^{\circ}+2x)+\cos (50^{\circ}-2x)]+[\cos (30^{\circ}-2x)+\cos (70^{\circ}-2x)]=0\iff$](//latex.artofproblemsolving.com/5/a/d/5adca794cf94ac89971734704dadac9a58832b8b.png)
See and here ==> http://www.artofproblemsolving.com/blog/32104
GENERALIZATION. Let be a convex quadrilateral with 
, 
,
and 
, where 
. Prove that 
.
PP5. Prove that
.
Proof. Denote
. Then for any 
we have 
.
In the particular case
obtain that 
and 
. Observe that 
, i.e.
. In conclusion, 
.
Remark. 
.
PP6. Show that
, where 
, 
.
Proof. If denote
, then 
are the roots of the equation 
,
i.e.
. Plugging 
obtain that 
, where
, where 
. Thus, 
.
Remark. A geometrical interpretation of this identity. Let
be a regular polygon inscribed in a circle with radius 
. Then 
.
The new identity
is a paricular case of the well-known identity 
for 
.
Indeed,
.
GENERALIZATION.
. Thus, 
is a continue function.
PP7. Prove that
, where .
Proof 1 (with complex numbers).
, 
and 
for any 
we have 
. Thus, 
.
Therefore, our identity is equivalently with an identity in![$\mathbb C[X]\ :\ \left(\frac {z^2+1}{2z}\right)^7+\left(\frac {w^2z^2+1}{2wz}\right)^7+\left(\frac {z^2+w^2}{2wz}\right)^7=\frac {63}{64}\cdot\frac {z^6+1}{2z^3}\iff$](//latex.artofproblemsolving.com/6/a/5/6a59424c844c4cc16a7f3e2695d613f98277b47a.png)
, what is truly because 
for any 
and .
Proof 2 (with fundamental symmetrical forms). Denote
. Prove easily that
![$\left\{\begin{array}{ccc}
s_1=a+(b+c)=\cos x+2\cos (x+\pi )\cos\frac {\pi}{3}=\cos x-\cos x & \implies & \boxed {s_1=0}\\\\
s_2=bc+a(b+c)=bc-a^2\implies 2s_2=\cos (2x+2\pi )+\cos\frac {2\pi}{3}-(1+\cos 2x) & \implies & \boxed{s_2=-\frac 34}\\\\
s_3=a(bc)\implies 4s_3=2\cos x\left[\cos (2x+2\pi )+\cos\frac {2\pi}{3}\right]=2\cos x\cos 2x-\cos x & \implies & \boxed{s_3=\frac 14\cdot\cos 3x}\end{array}\right\|$](//latex.artofproblemsolving.com/4/c/1/4c157623cb4aa7162cda2f0ba055da0bff34a473.png)
.
Therefore,
are the roots of the equation 
. Denote 
, where .
Show easily that for any we have 
, where 
, 
and 
. Thus,
.
Lemma. Consider the real numbers
, 
and 
, where 
.
Then
and 
.
Proof.
![$\sum_{k=1}^n\left[\sin \left(x_k+\frac r2\right)-\sin \left(x_k-\frac r2\right)\right]=$](//latex.artofproblemsolving.com/b/6/5/b65f20be3141c4350c488cff10faee926847fcd9.png)
, because 
. In conclusion, 
.
![$\sum_{k=1}^n\left[\cos \left(x_k-\frac r2\right)-\cos\left(x_k+\frac r2\right)\right]=$](//latex.artofproblemsolving.com/0/8/e/08e17816cd02bed1a2a2d3d7e4e15fe9ffe93956.png)
, because . In conclusion, 
.
PP8. Denote
, where 
and 
. Find 
as a function of 
.
Proof.
, where 
. Apply upper lemma for 
and
obtain that
. In conclusion, 
.
PP9. Prove that
and 
.
Proof. I"ll use the remarkable identity
. Indeed,
![$S_n=\sum_{k=1}^n\cos\left[\frac {k\pi}{2n+1}+(k-1)\pi\right]=$](//latex.artofproblemsolving.com/5/b/9/5b90a695e36f04c23e1c45edc7e7bd36e819349c.png)
because
and 
. Remark that 
.
Denote
. Observe that the roots of the equation 
are 
, whee 
, i.e. 
.
For
obtain that 
.
Remark. I used the evident relations
of the product/sum.
PP10. Ascertain
.
Proof 0 (trigonometric - pco's). Denote
. Observe that
. Thus, 
.
Remark. Let the arthmetrical progression
, i.e. 
, where 
(constant).
Prove easily that
and 
.
Proof 1 (trigonometric). Observe that
.
Proof 2 (geometric). Consider an
-isosceles 
with 
. Denote the point 
for which 
.
Observe that
. Let 
be the midpoints of the segments ![$[AB]\ ,[CD]$](//latex.artofproblemsolving.com/6/3/6/636c40633cc012b952b3ae9c9ae3d87869959f2a.png)
respectively. Thus, 
. Observe that 
and 
.
In conclusion,
.
Remark. Suppose w.l.o.g.
and denote 
. Thus, 
. From the relation 
obtain that
, i.e. 
and 
.
. So .
PP11. Prove that
is constant for any 
so that 
.
Proof 1. I"ll use the well-known conditioned identity
.
Proof 2. I"ll use the remarkable identity
. Thue, ![$E=\tan x\left[\tan (x+60)+\tan (x-60)\right]+\tan (x+60)\tan (x-60)=$](//latex.artofproblemsolving.com/e/b/2/eb2c9c3a91ade396916e3257d79719821c985c93.png)
![$\frac {\sin x}{\cos x}\cdot\frac {\sin [(x+60)+(x-60)]}{\cos (x+60)\cos (x-60)}+\frac {\sin (x+60)\sin (x-60)}{\cos (x+60)\cos (x-60)}=$](//latex.artofproblemsolving.com/8/9/f/89f46d2dba4c17e8cdb731626f605f6b4b8f62a2.png)
.
Proof 3 (common). Denote
. Thus, 
![$\frac {t\left[(t+\sqrt 3)(1+t\sqrt 3)+(1-t\sqrt 3)(t-\sqrt 3)\right]+\left(t^2-3\right)}{1-3t^2}=$](//latex.artofproblemsolving.com/d/2/7/d274fa53ed2594cad92fc634bf69339ee1548826.png)
.
PP12. Show that:
.
Proof.
, what is truly.
PP13. Ascertain
, where 
and for 
, find the limit 
.
Proof 1. I"ll use the well-known property of the symbol
, where 
. 
![$x\cdot \left[S_n(x)-nx^n\right]+\frac {x\cdot\left(1-x^n\right)}{1-x}\implies$](//latex.artofproblemsolving.com/d/9/d/d9d0f65e443b23b912a4c3dbb4ebc32b406d7244.png)
.
In conclusion,
and if denote 
, then obtain that 
.
Proof 2.
-\left(x^{n+1}-x\right)}{(x-1)^2}$](//latex.artofproblemsolving.com/b/7/9/b79defa9f6f6c23c684e97488a886b52a231c047.png)
![$\boxed{S_n(x)=\frac {x\cdot\left[nx^{n+1}-(n+1)x^n+1\right]}{(x-1)^2}}$](//latex.artofproblemsolving.com/5/c/c/5cc355fbc10e93375c459b68aa04a0ff2e856119.png)
.
Proof (3). 
.
Remark. Obtain (or with the Horner's schema) that![$\boxed{nx^{n+1}-(n+1)x^n+1=(x-1)^2\cdot\left[\ 1+2x+3x^2+\ \ldots\ +(n-1)x^{n-2}+nx^{n-1}\right]\ }\ (*)$](//latex.artofproblemsolving.com/0/0/3/0036183f7bb368186525db934324332e2af8462c.png)
and
![$\boxed{\ nx^n-x^{n-1}-x^{n-2}-\ \ldots\ -x^2-x-1=(x-1)\left[nx^{n-1}+(n-1)x^{n-2}+\ \ldots\ +3x^2+2x+1\right]\ }$](//latex.artofproblemsolving.com/e/d/c/edc288f91225c2aa1a011f5518d747b90c8d5b11.png)
.
From the relation obtain that 
, with the equality iff 
. Using the substitution ![$x:=\sqrt [n+1]a\ ,\ a>0$](//latex.artofproblemsolving.com/b/2/0/b207d0467cc68326476a647f5d63eeadc76075a9.png)
obtain that the inequality ![$\frac {1+na}{n+1}\ge \sqrt[n+1]{a^n}\ (1)$](//latex.artofproblemsolving.com/2/9/d/29da028dd57878247d8cb3fc53ff67878aa06c2e.png)
, where 
,
i.e. the A.M./G.M.-inequality in the case
for any 
and 
. Let 
be positive numbers for which denote 
, where 
is the arithmetical mean and 
,
where
is the geometrical mean. Using the inequality 
for 
obtain that ![$1+n\cdot\frac {G_n}{a_{n+1}}\ge (n+1)\cdot\sqrt [n+1]{\frac {G_n^n}{a_{n+1}^n}\cdot\frac {a_{n+1}}{a_{n+1}}}\iff$](//latex.artofproblemsolving.com/4/5/e/45e2c0f2feb084c84e36a814b8f4ea6b5cb78504.png)
(T. Popoviciu's inequality).
such that 
. Prove that 
.Proof. Are well-known the Viete's relations
. Denote 
, where 
. Observe that 
. Since 
obtain that
. In conclusion, 
.Particular case. For any
exists the inequality ![$(x+1)^2+x^2+1\ \ge\ 3\cdot\sqrt [3]{2x^2(x+1)^2}$](http://latex.artofproblemsolving.com/5/a/9/5a9986f31556e2f609507e3ba64e846b3f270e63.png)
.PP2. Prove that in a
triangle we have the relation 
.Proof 1.
.Remark. I used the well-known identity
which is equivalently with the identity 
.Indeed, from the relation
obtain that 
. I supposed well-known that 
.Proof 2.
.Remark. I used the well-known identity
. Indeed, 
.Proof 3.
.Proof 4.
.Remark.
. Define 
. Then 
.Indeed, the first relation becomes
and the second relation becomes
.PP3. Ascertain the monic polynomial equation which has the roots
.Proof. Using the remarkable sum
obtain that 
. Otherwise. 
.
![$-2\left[\left(\cos\frac {3\pi}{7}+\cos\frac {\pi}{7}\right)-\cos\frac {2\pi}{7}\right]\stackrel{(*)}{=}$](http://latex.artofproblemsolving.com/a/f/8/af8ab4edca0b57b1416fc1c6efb2666f28fec0c7.png)
![$-2\left[\left(\frac 12-\cos\frac {5\pi}{7}\right)+\cos\frac {2\pi}{7}\right]=$](http://latex.artofproblemsolving.com/d/8/4/d841fa8275f02ad8fb916d77db25798a4d742958.png)
. 
.In conclusion,
are the roots of the equation 
, i.e. 
.PP4. Let
be a convex quadrilater with 
and 
. Prove that 
.Proof 1 ("slicing"). . I"ll use the well-known lemma "Let
be a convex quadrilateral so that 
. Then 
" (you can prove easily).Denote
so that 
. Since 
and 
obtain from upper lemmathat
, i.e. 
. Since 
and 
obtain again fromupper lemma that
, i.e. 
. In conclusion, and 
, i.e. 
.Proof 2 ("slicing"). Let's draw equilateral triangle
,where the poiny 
lies on the same semi-plane with the point 
w.r.t. 
. Hence 
. Hence 
the streight line 
is the symmetric line of side 
. Then 
. Let
. Hence 
. The symmetricline of the side
will pass throught the points 
and . 
and hence 
. Let's drawa line dividing
into two equal parts and let's this line crosses the straight line 
in point 
. Hence 
.Then
will be an isosceles triangle 
. 
and the triangle 
will be an isosceles triangle.
, but at the same time and 
. Then 
and
. Let 
. Then in 
.Proof 3 (metric).
quadrilater is orthodiagonal if and only if 
. I"ll use the Sinus' theorem :
.Replacing in
the last 4 relations we only have to prove that 
.
.I used the well-known identities
.Proof 4 (trigonometric - brute force). Denote
. Thus, 
. Apply an well-known identity
. Therefpre,
![$\cos 40^{\circ}[\cos 10^{\circ}-\cos (10^{\circ}+2x)]=\cos 10^{\circ}[\sin 50^{\circ}+\sin (30^{\circ}+2x)]\iff$](http://latex.artofproblemsolving.com/3/e/0/3e086bacb8d55ccb54649f01bdaf21b97463f303.png)
Remark. From the step
can continue thus :
![$ [\cos (50^{\circ}+2x)+\cos (50^{\circ}-2x)]+[\cos (30^{\circ}-2x)+\cos (70^{\circ}-2x)]=0\iff$](http://latex.artofproblemsolving.com/5/a/d/5adca794cf94ac89971734704dadac9a58832b8b.png)
See and here ==> http://www.artofproblemsolving.com/blog/32104
GENERALIZATION. Let
be a convex quadrilateral with 
, 
,
and 
, where 
. Prove that 
.PP5. Prove that
.Proof. Denote
. Then for any 
we have 
.In the particular case
obtain that 
and 
. Observe that 
, i.e.
. In conclusion, 
.Remark.
.PP6. Show that
, where 
, 
.Proof. If denote
, then 
are the roots of the equation 
,i.e.
. Plugging 
obtain that 
, where
, where 
. Thus, 
.Remark. A geometrical interpretation of this identity. Let
be a regular polygon inscribed in a circle with radius 
. Then 
.The new identity
is a paricular case of the well-known identity 
for 
.Indeed,
.GENERALIZATION.
. Thus, 
is a continue function.PP7. Prove that
, where .Proof 1 (with complex numbers).
, 
and 
for any 
we have 
. Thus, 
.Therefore, our identity is equivalently with an identity in
![$\mathbb C[X]\ :\ \left(\frac {z^2+1}{2z}\right)^7+\left(\frac {w^2z^2+1}{2wz}\right)^7+\left(\frac {z^2+w^2}{2wz}\right)^7=\frac {63}{64}\cdot\frac {z^6+1}{2z^3}\iff$](http://latex.artofproblemsolving.com/6/a/5/6a59424c844c4cc16a7f3e2695d613f98277b47a.png)
, what is truly because 
for any 
and .Proof 2 (with fundamental symmetrical forms). Denote
. Prove easily that![$\left\{\begin{array}{ccc}
s_1=a+(b+c)=\cos x+2\cos (x+\pi )\cos\frac {\pi}{3}=\cos x-\cos x & \implies & \boxed {s_1=0}\\\\
s_2=bc+a(b+c)=bc-a^2\implies 2s_2=\cos (2x+2\pi )+\cos\frac {2\pi}{3}-(1+\cos 2x) & \implies & \boxed{s_2=-\frac 34}\\\\
s_3=a(bc)\implies 4s_3=2\cos x\left[\cos (2x+2\pi )+\cos\frac {2\pi}{3}\right]=2\cos x\cos 2x-\cos x & \implies & \boxed{s_3=\frac 14\cdot\cos 3x}\end{array}\right\|$](http://latex.artofproblemsolving.com/4/c/1/4c157623cb4aa7162cda2f0ba055da0bff34a473.png)
.Therefore,
are the roots of the equation 
. Denote 
, where .Show easily that for any
we have 
, where 
, 
and 
. Thus,
.Lemma. Consider the real numbers
, 
and 
, where 
.Then
and 
.Proof.
![$\sum_{k=1}^n\left[\sin \left(x_k+\frac r2\right)-\sin \left(x_k-\frac r2\right)\right]=$](http://latex.artofproblemsolving.com/b/6/5/b65f20be3141c4350c488cff10faee926847fcd9.png)
, because 
. In conclusion, 
.
![$\sum_{k=1}^n\left[\cos \left(x_k-\frac r2\right)-\cos\left(x_k+\frac r2\right)\right]=$](http://latex.artofproblemsolving.com/0/8/e/08e17816cd02bed1a2a2d3d7e4e15fe9ffe93956.png)
, because . In conclusion, 
.PP8. Denote
, where 
and 
. Find 
as a function of 
.Proof.
, where 
. Apply upper lemma for 
andobtain that
. In conclusion, 
.PP9. Prove that
and 
.Proof. I"ll use the remarkable identity
. Indeed,![$S_n=\sum_{k=1}^n\cos\left[\frac {k\pi}{2n+1}+(k-1)\pi\right]=$](http://latex.artofproblemsolving.com/5/b/9/5b90a695e36f04c23e1c45edc7e7bd36e819349c.png)
because
and 
. Remark that 
.Denote
. Observe that the roots of the equation 
are 
, whee 
, i.e. 
.For
obtain that 
.Remark. I used the evident relations
of the product/sum.PP10. Ascertain
.Proof 0 (trigonometric - pco's). Denote
. Observe that
. Thus, 
.Remark. Let the arthmetrical progression
, i.e. 
, where 
(constant).Prove easily that
and 
.Proof 1 (trigonometric). Observe that
.Proof 2 (geometric). Consider an
-isosceles 
with 
. Denote the point 
for which 
.Observe that
. Let 
be the midpoints of the segments ![$[AB]\ ,[CD]$](http://latex.artofproblemsolving.com/6/3/6/636c40633cc012b952b3ae9c9ae3d87869959f2a.png)
respectively. Thus, 
. Observe that 
and 
.In conclusion,
.Remark. Suppose w.l.o.g.
and denote 
. Thus, 
. From the relation 
obtain that
, i.e. 
and 
.
. So .PP11. Prove that
is constant for any 
so that 
.Proof 1. I"ll use the well-known conditioned identity
.Proof 2. I"ll use the remarkable identity
. Thue, ![$E=\tan x\left[\tan (x+60)+\tan (x-60)\right]+\tan (x+60)\tan (x-60)=$](http://latex.artofproblemsolving.com/e/b/2/eb2c9c3a91ade396916e3257d79719821c985c93.png)
![$\frac {\sin x}{\cos x}\cdot\frac {\sin [(x+60)+(x-60)]}{\cos (x+60)\cos (x-60)}+\frac {\sin (x+60)\sin (x-60)}{\cos (x+60)\cos (x-60)}=$](http://latex.artofproblemsolving.com/8/9/f/89f46d2dba4c17e8cdb731626f605f6b4b8f62a2.png)
.Proof 3 (common). Denote
. Thus, 
![$\frac {t\left[(t+\sqrt 3)(1+t\sqrt 3)+(1-t\sqrt 3)(t-\sqrt 3)\right]+\left(t^2-3\right)}{1-3t^2}=$](http://latex.artofproblemsolving.com/d/2/7/d274fa53ed2594cad92fc634bf69339ee1548826.png)
.PP12. Show that:
.Proof.
, what is truly. PP13. Ascertain
, where 
and for 
, find the limit 
.Proof 1. I"ll use the well-known property of the symbol
, where 
. 
![$x\cdot \left[S_n(x)-nx^n\right]+\frac {x\cdot\left(1-x^n\right)}{1-x}\implies$](http://latex.artofproblemsolving.com/d/9/d/d9d0f65e443b23b912a4c3dbb4ebc32b406d7244.png)
.In conclusion,
and if denote 
, then obtain that 
.Proof 2.
-\left(x^{n+1}-x\right)}{(x-1)^2}$](http://latex.artofproblemsolving.com/b/7/9/b79defa9f6f6c23c684e97488a886b52a231c047.png)
![$\boxed{S_n(x)=\frac {x\cdot\left[nx^{n+1}-(n+1)x^n+1\right]}{(x-1)^2}}$](http://latex.artofproblemsolving.com/5/c/c/5cc355fbc10e93375c459b68aa04a0ff2e856119.png)
.Proof (3).
.Remark. Obtain (or with the Horner's schema) that
![$\boxed{nx^{n+1}-(n+1)x^n+1=(x-1)^2\cdot\left[\ 1+2x+3x^2+\ \ldots\ +(n-1)x^{n-2}+nx^{n-1}\right]\ }\ (*)$](http://latex.artofproblemsolving.com/0/0/3/0036183f7bb368186525db934324332e2af8462c.png)
and
![$\boxed{\ nx^n-x^{n-1}-x^{n-2}-\ \ldots\ -x^2-x-1=(x-1)\left[nx^{n-1}+(n-1)x^{n-2}+\ \ldots\ +3x^2+2x+1\right]\ }$](http://latex.artofproblemsolving.com/e/d/c/edc288f91225c2aa1a011f5518d747b90c8d5b11.png)
.From the relation
obtain that 
, with the equality iff 
. Using the substitution ![$x:=\sqrt [n+1]a\ ,\ a>0$](http://latex.artofproblemsolving.com/b/2/0/b207d0467cc68326476a647f5d63eeadc76075a9.png)
obtain that the inequality ![$\frac {1+na}{n+1}\ge \sqrt[n+1]{a^n}\ (1)$](http://latex.artofproblemsolving.com/2/9/d/29da028dd57878247d8cb3fc53ff67878aa06c2e.png)
, where 
,i.e. the A.M./G.M.-inequality in the case
for any 
and 
. Let 
be positive numbers for which denote 
, where 
is the arithmetical mean and 
,where
is the geometrical mean. Using the inequality 
for 
obtain that ![$1+n\cdot\frac {G_n}{a_{n+1}}\ge (n+1)\cdot\sqrt [n+1]{\frac {G_n^n}{a_{n+1}^n}\cdot\frac {a_{n+1}}{a_{n+1}}}\iff$](http://latex.artofproblemsolving.com/4/5/e/45e2c0f2feb084c84e36a814b8f4ea6b5cb78504.png)
(T. Popoviciu's inequality). This post has been edited 140 times. Last edited by Virgil Nicula, Nov 21, 2015, 2:53 PM
August 2018
January 2018