Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

According to a paper by Zhiqin Lu in the Mathematical Gazette (the British publication, not the Boston-area newsletter, if that still exists (or even if it doesn't)) in 2007(?), if $u+v+w=\pi$ and $a,b,c \ge 0$, then $a + b + c \ge 2 \sqrt{bc} \cos u + 2 \sqrt{ca} \cos v + 2 \sqrt{ab} \cos w$. It wasn't hard to show that equality holds iff $a:b:c = \sin^2 u : \sin^2 v : \sin^2 w$, but if I recall correctly, that wasn't in the paper. I would think that there must be some natural geometric interpretation of this proposition. What is it?

The case where equality holds says that if $u+v+w = \pi$, then $\sin^2 u + \sin^2 v + \sin^2 w = 2 \sin u \sin v \cos w + 2 \sin u \cos v \sin w + 2 \cos u \sin v \sin w$. That one has a simple geometric interpretation as a sort of mash-up of the law of sines and the law of cosines. But now suppose we say that if $$\sum_i u_i =\pi$$ then $$ \sum_{i=1}^\infty \sin^2 u_i = \sum_{\text{even }n\ge 2} (-1)^{(n-2)/2}n\sum_{|A|=n} \prod_{i\in A}\sin u_i \prod_{i\not\in A} \cos u_i $$ The case of a sum of three variables that add up to $\pi$ is the special case in which all but three of these are 0. Does it have a geometric meaning?

Later edit: So far we have an answer about the inequality, but not yet about the equality. I will probably comment soon on the former.

share|cite|improve this question
Michael: the Boston-area Mathematical Gazette still exists. I got an email about the latest issue last week. You can check it out online at –  KConrad Apr 25 '11 at 5:16
I have posted a proof of the inequality on MathLinks some years ago, and IIRC it reduced it to a $\left(\text{some distance}\right)^2\geq 0$. Can anyone find it? –  darij grinberg Apr 25 '11 at 10:00
Found it. See my answer below. –  darij grinberg Apr 25 '11 at 10:21
To say that the square of some particular distance is nonnegative tells us only what we would know without knowing which particular distance it is. –  Michael Hardy Apr 25 '11 at 19:12

1 Answer 1

In an ancient MathLinks topic (post #6; but see below for a copy) I have given a proof of the inequality by reducing it to $\left(\sqrt{a}\vec{p}+\sqrt{b}\vec{q}+\sqrt{c}\vec{r}\right)^2\geq 0$, where multiplication means scalar product of vectors and $\vec{p}$, $\vec{q}$, $\vec{r}$ are unit length vectors chosen in such a way that the angles between them are $\pi-u$, $\pi-v$, $\pi-w$, respectively. This rewrites geometrically as follows: Pick a point $P$ in the plane, and take three points $A$, $B$, $C$ such that $PA=\sqrt{a}$, $PB=\sqrt{b}$, $PC=\sqrt{c}$, $\measuredangle BPC=\pi-u$, $\measuredangle CPA=\pi-v$ and $\measuredangle APB=\pi-w$. Then, the difference between the left hand side and the right hand side of your inequality is $9$ times the square of the distance between the point $P$ and the centroid of triangle $ABC$. Equality thus holds if and only if $P$ is the centroid of triangle $ABC$; this is equivalent to the assertion that the triangles $BPC$, $CPA$, $APB$ have equal areas; this, in turn, is equivalent to the assertion that $\sqrt{a}:\sqrt{b}:\sqrt{c}=\sin u:\sin v:\sin w$ (because the area of triangle $BPC$ is $\frac{1}{2}\cdot PB\cdot PC\cdot \sin\measuredangle BPC=\frac{1}{2}\sqrt{b}\sqrt{c}\sin u$ etc.).

For better searchability, let me copy my MathLinks posts over here (finding some old post on MathLinks is almost impossible as for now). Note that I do not claim originality for the theorems.

Theorem 1. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C = 180^{\circ}$. Then,

$x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$.

Proof of Theorem 1. We will denote by $\measuredangle\left(\overrightarrow{p};\;\overrightarrow{q}\right)$ the directed angle between two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ (note that this is a directed angle modulo $360^{\circ}$).

For any two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$, we are going to denote by $\overrightarrow{p}\cdot\overrightarrow{q}$ the scalar product of the vectors $\overrightarrow{p}$ and $\overrightarrow{q}$.

For any vector $\overrightarrow{p}$, we are going to denote by $\overrightarrow{p}^2$ the scalar product $\overrightarrow{p}\cdot\overrightarrow{p}$. Every vector $\overrightarrow{p}$ satisfies $\overrightarrow{p}^2=\left|\left|\overrightarrow{p}\right|\right|^2\geq 0$.

Let $\overrightarrow{a}$ be a vector of unit length. Let $\overrightarrow{b}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)=180^{\circ}-C$. Let $\overrightarrow{c}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$. Then,

$\measuredangle\left(\overrightarrow{c};\;\overrightarrow{a}\right)=360^{\circ}-\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)-\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)$ $=360^{\circ}-\left(180^{\circ}-C\right)-\left(180^{\circ}-A\right)=C+A=180^{\circ}-B$

(since A + B + C = 180°).

Now, all the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ have unit length: $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$. Thus, $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$ yields

$\overrightarrow{b}\cdot\overrightarrow{c}=\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|\cdot\cos\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=1\cdot 1\cdot\cos\left(180^{\circ}-A\right)=\cos\left(180^{\circ}-A\right)$ $=-\cos A$.

Similarly, we obtain $\overrightarrow{c}\cdot\overrightarrow{a}=-\cos B$ and $\overrightarrow{a}\cdot\overrightarrow{b}=-\cos C$. Thus,



$+2\cdot y\cdot\overrightarrow{b}\cdot z\cdot\overrightarrow{c}+2\cdot z\cdot\overrightarrow{c}\cdot x\cdot\overrightarrow{a}+2\cdot x\cdot\overrightarrow{a}\cdot y\cdot\overrightarrow{b}$

$=x^2\underbrace{\cdot\left|\overrightarrow{a}\right|^2}_{=1^2}+y^2\cdot\underbrace{\left|\overrightarrow{b}\right|^2}_{=1^2}+z^2\cdot\underbrace{\left|\overrightarrow{c}\right|^2}_{=1^2}+2yz\cdot\underbrace{\overrightarrow{b}\cdot\overrightarrow{c}}_{=-\cos A}+2zx\cdot\underbrace{\overrightarrow{c}\cdot\overrightarrow{a}}_{=-\cos B}+2xy\cdot\underbrace{\overrightarrow{a}\cdot\overrightarrow{b}}_{=-\cos C}$

$=x^2\cdot 1^2+y^2\cdot 1^2+z^2\cdot 1^2+2yz\cdot\left(-\cos A\right)+2zx\cdot\left(-\cos B\right)+2xy\cdot\left(-\cos C\right)$

$=x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C$.

Since we, obviously, have $\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2\geq 0$, we thus get $x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C\geq 0$, so that $x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$, and Theorem 1 is proven.

Other proofs of Theorem 1 can be found at and .

Theorem 1 is equivalent to the following, also quite useful (for olympiad mathematics and magazine problem sections, that is, although I would not be surprised to see more applications) inequality:

Theorem 2. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C$ is a multiple of $180^{\circ}$. Then,

$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$.

We will only show a proof of Theorem 2 using Theorem 1: First, we can WLOG assume that $A + B + C = 180^{\circ}$. This is because the inequality

$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$

will not change if we add a multiple of 180° to one of the angles A, B and C (because $ \sin^{2}\left(180^{\circ} + u\right) = \sin^{2}u$ for every u), and consequently, since A + B + C is a multiple of 180°, we can add a multiple of 180° to the angle A such that, after this, we will have A + B + C = 180°.

Now, for A + B + C = 180°, we have

$ \left(180^{\circ} - 2A\right) + \left(180^{\circ} - 2B\right) + \left(180^{\circ} - 2C\right) = 540^{\circ} - 2\cdot\left(A + B + C\right)$ $ = 540^{\circ} - 2\cdot 180^{\circ} = 180^{\circ}$.

Hence, Theorem 1 (applied to $ 180^{\circ}-2A$, $ 180^{\circ}-2B$, $ 180^{\circ}-2C$ instead of $ A$, $ B$, $ C$) yields

$ x^{2} + y^{2} + z^{2}\geq 2yz\cos\left(180^{\circ} - 2A\right) + 2zx\cos\left(180^{\circ} - 2B\right) + 2xy\cos\left(180^{\circ} - 2C\right)$.

Since $ \cos\left(180^{\circ} - 2A\right) = - \cos\left(2A\right) = - \left(1 - 2\sin^{2}A\right) = 2\sin^{2}A - 1$ and similarly $ \cos\left(180^{\circ} - 2B\right) = 2\sin^{2}B - 1$ and $ \cos\left(180^{\circ} - 2C\right) = 2\sin^{2}C - 1$, this becomes

$ x^{2} + y^{2} + z^{2}\geq 2yz\left(2\sin^{2}A - 1\right) + 2zx\left(2\sin^{2}B - 1\right) + 2xy\left(2\sin^{2}C - 1\right)$ $ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right) - \left(2yz + 2zx + 2xy\right)$ $ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2} + \left(2yz + 2zx + 2xy\right)\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$ $ \Longleftrightarrow\ \ \ \ \ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$,

and Theorem 2 is proven.

Theorem 2 also trivially follows from and was also discussed at ...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.