# Oppenheim's Inequality for triangles, American Mathematical Monthly problems

I did a proof for the inequality below, and I would like know if anyone also has a trigonometric proof for this inequality.If you have a trigonometric demonstration, please post your solution. This problem appeared in the American Mathematical Monthly magazine in 1965, the inequality was proposed in that form by Sir Alexander Oppenheim:

Let $x,y,z$ positive real numbers and $\Delta ABC$ a triangle. $\displaystyle [ABC]$ denotes the triangle area and $\displaystyle a,b,c$ the sides of the triangle. The inequality below is true: $$a^2x+b^2y+c^2z\geq 4[ABC]\sqrt{xy+xz+yz}$$

Various inequalities can be deduced through this inequality, for example, Weitzenböck's inequality, Neuberg-Pedoe inequality, Hadwiger-Finsler inequality, and so on. I'll post my solution right below. $$Proof$$

Let $\alpha,\beta,\gamma$ denote the opposite angles to the sides $a, b, c$, respectively. $R$ is the circumradius of $\Delta ABC$. Observe that: $$a^2x+b^2y+c^2z\geq 4[ABC]\sqrt{xy+xz+yz}$$ $$a^2x+b^2y+c^2z\geq \frac{abc}{R}\sqrt{xy+xz+yz}$$ $$\frac{aRx}{bc}+\frac{bRy}{ac}+\frac{cRz}{ab}\geq \sqrt{xy+xz+yz}$$ $$\frac{1}{2}\left(\frac{4aR^2x}{2Rbc}+\frac{4bR^2y}{2Rac}+\frac{4cR^2z}{2Rab}\right)\geq \sqrt{xy+xz+yz}$$ $$x\frac{\sin\alpha }{\sin\beta \sin \gamma}+y\frac{\sin\beta }{\sin\alpha \sin \gamma}+z\frac{\sin\gamma }{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

$$x\frac{\sin(\pi-\alpha) }{\sin\beta \sin \gamma}+y\frac{\sin(\pi-\beta )}{\sin\alpha \sin \gamma}+z\frac{\sin(\pi-\gamma )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

$$x\frac{\sin(\alpha+\beta+\gamma-\alpha) }{\sin\beta \sin \gamma}+y\frac{\sin(\alpha+\beta+\gamma-\beta )}{\sin\alpha \sin \gamma}+z\frac{\sin(\alpha+\beta+\gamma-\gamma )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

$$x\frac{\sin(\beta+\gamma) }{\sin\beta \sin \gamma}+y\frac{\sin(\alpha+\gamma )}{\sin\alpha \sin \gamma}+z\frac{\sin(\alpha+\beta )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

$$x\frac{(\sin\beta \cos\gamma+\sin\gamma \cos \beta) }{\sin\beta \sin \gamma}+y\frac{(\sin\alpha \cos\gamma+\sin\gamma \cos \alpha) }{\sin\alpha \sin \gamma}+z\frac{(\sin\alpha \cos\beta+\sin\beta \cos \alpha) }{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

$$(\cot\beta+\cot\gamma)x+(\cot\alpha+\cot\gamma)y+(\cot\alpha+\cot\beta)z\geq 2\sqrt{xy+xz+yz} \tag{1}$$

Since inequality is homogeneous in the variables $x,y,z$, do it $\displaystyle xy+xz+yz=1$ and take the substitution $\displaystyle x=\cot\alpha',y=\cot\beta',z=\cot\gamma'$, we have que $\displaystyle \alpha',\beta',\gamma'$ are angles of a triangle, and our inequality will be equivalent to the inequality below:

$$(\cot\beta+\cot\gamma)\cot\alpha'+(\cot\alpha+\cot\gamma)\cot\beta'+(\cot\alpha+\cot\beta)\cot\gamma'\geq 2 \tag{2}$$ Suppose without loss of generality that (the reverse case is analogous) :

$$\cot\alpha \geq \cot \alpha' \tag{3}$$ $$\cot\beta \geq \cot \beta' \tag{4}$$ $$\cot\gamma'\geq \cot \gamma \tag{5}$$ Because these variables are angles of a triangle, we can not have $\cot \alpha \geq \cot \alpha' , \cot\beta \geq \cot \beta', \cot \gamma\geq \cot\gamma'$.In fact, this can not occur, since it supposes without loss of generality that $\displaystyle \alpha'\geq\alpha$ and $\displaystyle \beta'\geq\beta$(as the cotangent is decreasing, this implies that $\displaystyle \cot\alpha \geq \cot \alpha'$ and $\displaystyle \cot\beta \geq \cot \beta'$), summing up these first two inequalities we have:

$\\ \\ \displaystyle \alpha'+\beta'\geq \alpha+\beta \Rightarrow \cot(\alpha+\beta)\geq \cot(\alpha'+\beta')\Rightarrow -\cot(\pi-\alpha+\beta)\geq- \cot(\pi-\alpha'+\beta') \Rightarrow -\cot(\alpha+\beta+\gamma-(\alpha+\beta))\geq- \cot(\alpha'+\beta'+\gamma'-(\alpha'+\beta')) \Rightarrow -\cot(\gamma)\geq- \cot(\gamma') \Rightarrow \cot(\gamma')\geq \cot(\gamma)\\ \\$

Now set the $\displaystyle f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma'):\mathbb{R}^6\rightarrow \mathbb{R}$ and $\displaystyle f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma'):\mathbb{R}^6\rightarrow \mathbb{R}$ such that:

\begin{equation*} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')= \end{equation*} $$(\cot\beta+\cot\gamma)(\cot\alpha'-\cot\alpha)+(\cot\alpha+\cot\gamma)(\cot\beta'-\cot\beta)+(\cot\alpha+\cot\beta)(\cot\gamma'-\cot\gamma) \tag{6}$$

\begin{equation*} f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')= \end{equation*} $$(\cot\beta'+\cot\gamma')(\cot\alpha-\cot\alpha')+(\cot\alpha'+\cot\gamma')(\cot\beta-\cot\beta')+(\cot\alpha'+\cot\beta')(\cot\gamma-\cot\gamma') \tag{7}$$

Note now that by inequalities (3), (4) and (5) it follows that:

$$0 \geq \cot\alpha'-\cot\alpha \tag{8}$$

$$0 \geq \cot \beta' -\cot\beta \tag{9}$$

$$\cot\gamma'-\cot\gamma \geq 0 \tag{10}$$ We know that $\displaystyle \alpha',\beta',\gamma'$ are angles of a triangle, so there exists $\displaystyle a',b',c'$ such that $\displaystyle a'^2=b'^2+c'^2-2b'c'\cos\alpha',b'^2=a'^2+c'^2-2a'c'\cos\beta',c'^2=a'^2+b'^2-2a'b'\cos\gamma'$.Let $\displaystyle R'$ the circumradius of the triangle of sides $\displaystyle a',b',c'$.

Let

$\displaystyle k_{\alpha',\beta',\gamma'}:=\frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}$, therefore:

$$\frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}=k_{\alpha',\beta',\gamma'} \tag{11}$$

Where $\displaystyle k_{\alpha',\beta',\gamma'}$ is a real variable of any kind.And since our original inequality is homogeneous in the variables a, b, c, suppose without loss of generality that the equality below occurs:

$$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=k_{\alpha',\beta',\gamma'} \tag{12}$$

For each real value of fixed $\displaystyle k_{\alpha',\beta',\gamma'}$.Since x, y, z do not depend of the circumradius R ', suppose that $\displaystyle R'\geq R$.Take the inequality (3) and consider the development (applying the law of cosines and law of sines): $\\ \displaystyle \cot\alpha \geq \cot\alpha' \Rightarrow \frac{(b^2+c^2-a^2)R}{abc} \geq \frac{(b'^2+c'^2-a'^2)R'}{a'b'c'} \Rightarrow \left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}-2\frac{a}{bc}\right)R\geq \left(\frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}-2\frac{a'}{b'c'}\right)R' \Rightarrow Rk_{\alpha',\beta',\gamma'}-2\frac{aR}{bc}\geq Rk_{\alpha',\beta',\gamma'}-2\frac{a'R'}{b'c'} \Rightarrow \frac{a'R'}{b'c'}\geq \frac{aR}{bc} \Rightarrow$

$$\frac{a'R'}{b'c'}\geq \frac{aR}{bc} \tag{13}$$

Applying the same rationale for inequality (4), we conclude: $$\frac{b'R'}{a'c'}\geq \frac{bR}{ac} \tag{14}$$

Suppose by contradiction that it occurs:

$$\cot\alpha+\cot\gamma> \cot\alpha'+\cot\gamma' \tag{15}$$

See that:

$\\ \displaystyle \cot\alpha+\cot\gamma> \cot\alpha'+\cot\gamma' \Rightarrow \frac{(b^2+c^2-a^2)R}{abc}+\frac{(a^2+b^2-c^2)R}{abc}> \frac{(b'^2+c'^2-a'^2)R'}{a'b'c'}+\frac{(a'^2+b'^2-c'^2)R'}{a'b'c'} \Rightarrow \frac{bR}{ac}>\frac{b'R'}{a'c'} \\$

This contradicts the inequality (14). On the other hand, suppose by contradiction that it occurs:

$$\cot\beta+\cot\gamma> \cot\beta'+\cot\gamma' \tag{16}$$

See that:

$\\ \displaystyle \cot\beta+\cot\gamma> \cot\beta'+\cot\gamma' \Rightarrow \frac{(a^2+c^2-b^2)R}{abc}+\frac{(a^2+b^2-c^2)R}{abc}> \frac{(a'^2+c'^2-b'^2)R'}{a'b'c'}+\frac{(a'^2+b'^2-c'^2)R'}{a'b'c'} \Rightarrow \frac{aR}{bc}>\frac{a'R'}{b'c'} \\$

$$\cot\alpha+\cot\gamma \leq \cot\alpha'+\cot\gamma' \tag{17}$$

$$\cot\beta+\cot\gamma \leq \cot\beta'+\cot\gamma' \tag{18}$$

Multiplying (17) by $\displaystyle \cot\beta'-\cot\beta$ and multiplying (18) by $\displaystyle \cot\alpha'-\cot\alpha$, note that these inequalities will reverse, since we are multiplying by non-positive quantities, we will have, respectively:

$$(\cot\alpha+\cot\gamma) (\cot\beta'-\cot\beta)\geq (\cot\alpha'+\cot\gamma')(\cot\beta'-\cot\beta) \tag{19}$$

$$(\cot\beta+\cot\gamma) (\cot\alpha'-\cot\alpha)\geq (\cot\beta'+\cot\gamma')(\cot\alpha'-\cot\alpha) \tag{20}$$

On the other hand of inequalities (3) and (4) we know that: $$\cot\alpha+\cot\beta \geq \cot\alpha'+\cot\beta' \tag{21}$$ Multiplying the above inequality by $\displaystyle \cot\gamma'-\cot\gamma$, that by the inequality (10) we know to be greater than or equal to zero, we will have:

$$(\cot\alpha+\cot\beta) (\cot\gamma'-\cot\gamma)\geq (\cot\alpha'+\cot\beta')(\cot\gamma'-\cot\gamma) \tag{22}$$ Adding (19), (20) and (22), we will have:

\begin{equation*} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq \end{equation*} $$(\cot\alpha'+\cot\gamma')(\cot\beta'-\cot\beta)+(\cot\beta'+\cot\gamma')(\cot\alpha'-\cot\alpha)+(\cot\alpha'+\cot\beta')(\cot\gamma'-\cot\gamma) \tag{23}$$

Adding the LHS of (23) with the LHS of (7) and the RHS of (23) with the RHS of (7), the terms will cancel and we will have:

$$f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')+f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq 0$$ And this implies, finally, that:

$$(\cot\beta+\cot\gamma)\cot\alpha'+(\cot\alpha+\cot\gamma)\cot\beta'+(\cot\alpha+\cot\beta)\cot\gamma'\geq 2$$ That is precisely the inequality (2), which is equivalent to the desired inequality.Thus, the inequality yields.

• What do you mean by "trigonometric proof" if you don't count this as trigonometric? It is hard to tell what the point of including your (long) proof is. Dec 26, 2016 at 3:05
• A proof that does not use algebraic inequalities... Dec 26, 2016 at 3:09
• What a work since your recent post ! Dec 26, 2016 at 3:44
• @Israel Meireles Chrisostomo There is a smooth and an easy algebraic proof. Dec 26, 2016 at 7:17
• @Michael Rozenberg,The problem of algebraic proof's is the restriction of on sign variable, if you take a trigonometric proof you can see that at least one variable can be negative, since xy+xz+yz are positive. Dec 26, 2016 at 7:26

Here is my algebraic proof.

We need to prove that:

$$(a^2x+b^2y+c^2z)^2\geq\sum\limits_{cyc}(2a^2b^2-a^4)(xy+xz+yz)$$ or $$c^4z^2-\left(\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2a^2c^2\right)x+\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2b^2c^2\right)y\right)z+$$ $$+a^4x^2+b^4y^2-\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2a^2b^2\right)xy\geq0,$$ for which it's enough to prove that $$\left(\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2a^2c^2\right)x+\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2b^2c^2\right)y\right)^2-$$ $$-4c^4\left(a^4x^2+b^4y^2-\left(\sum\limits_{cyc}(2a^2b^2-a^4)-2a^2b^2\right)xy\right)\leq0$$ or $$\sum\limits_{cyc}(2a^2b^2-a^4)\left((a^2+c^2-b^2)x-(b^2+c^2-a^2)y\right)^2\geq0.$$ Done!

• I am assuming the last inequality follows from some SOS or SOS-Schur but it does not seem trivial. Aug 31, 2022 at 1:58

A "sloppy" but quite straightforward proof. If someone comes up with a simple argument to solve the two points at the end I would be happy!

Writing the expression as $$\frac{a^2x+b^2y+c^2z}{\sqrt{xy+yz+zx}}\geq 4[ABC]$$ the RHS is independent of $x,y,z$, therefore it is necessary and sufficient to prove the inequality in the worst possible case, i.e. when the LHS is minimized in $x,y,z$.

In particular, by homogeneity we can fix $a,b,c$ and consider the problem $$\min\{a^2x+b^2y+c^2z:xy+yz+zx=1,\,x,y,z\geq0\}.$$

If two of $x,y,z$ are zero the inequality is trivially proven. If just one of them is zero, say $z$, then the problem becomes $$\min \{a^2x+b^2y:xy=1,\,x,y\geq 0\}=2ab$$ by AM-GM, and clearly $2ab\geq 2ab\sin\gamma=4[ABC]$.

The last case is $x,y,z>0$. By the Lagrange method we obtain a critical point in the interior where $$(a^2,b^2,c^2)=\lambda(y+z,z+x,x+y)$$ that is \begin{align} &x=b^2+c^2-a^2\\ &y=c^2+a^2-b^2\\ &z=a^2+b^2-c^2 \end{align} up to a multiplicative constant. Substituting above we obtain $$a^2x+b^2y+c^2z=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)=16[ABC]^2$$ by Heron's formula, and $$\sqrt{xy+yz+zx}=\sqrt{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)}=4[ABC]$$ again by Heron. In particular, in the critical point the equality holds.

The sloppyness comes from the fact that:

1) we don't know the critical point is actually a minimum (or do we?)

2) the infimum could be at infinity, think of $(x,y,z)=(\epsilon,\epsilon,\frac{1-\epsilon^2}{2\epsilon})$ with $\epsilon$ small

• yes - you need to look at the Hessian which could tell you are at a minimum (since the problem is true it will indeed tell you so). But this is only a local minimum. Good thing is we are in a bounded domain, so you simply need to check that there is no smaller value on the boundary of the domain. Aug 31, 2022 at 2:02

Given two triangles $\triangle A_1B_1C_1$, $\triangle A_2B_2C_2$ and positive numbers $x,y,z$.

Let $a_i,b_i,c_i$; $A_i, B_i, C_i$ and $\Delta_i$ be the sides, angles and area of triangle $\triangle A_iB_iC_i$.

We are going to show${}^{\color{blue}{[1]}}$ a two-triangle version of inequality in question. $$\bbox[padding: 1em;border:1px solid blue]{x a_1a_2 + y b_1 b_2 + z c_1 c_2 \ge 4\sqrt{(xy+yz+zx) \Delta_1 \Delta_2}}\tag{*1}$$

Let $u = xa_1a_2$, $v = yb_1b_2$, $w = zc_1c_2$ and $A_{\pm} = A_1 \pm A_2$, $B_{\pm} = B_1 \pm B_2$, $C_{\pm} = C_1 \pm C_2$.
Notice

$$2\Delta_1 = b_1c_1\sin A_1 = c_1a_1\sin B_1 = a_1b_1\sin C_1\\ 2\Delta_2 = b_2c_2\sin A_2 = c_2a_2\sin B_2 = a_2b_2\sin C_2$$ We have \begin{align} {\rm LHS}^2 - {\rm RHS}^2 &= (u + v + w)^2 - 4(uv\sin C_1\sin C_2 + vw \sin A_1\sin A_2 + wu \sin B_1\sin B_2)\\ &= u^2 + v^2 + w^2 + 2(uvW + vwU + wuV ) \end{align} where $U = 1 - 2\sin A_1\sin A_2$, $V = 1 - 2\sin B_1\sin B_2$ and $W = 1 - 2\sin C_1 \sin C_2$. Notice $$U = 1 - 2\sin A_1\sin A_2 = 1 + \cos A_+ - \cos A_- = \cos A_+ + \frac12 \sin^2 \frac{A_-}{2} \ge \cos A_+$$ and similar inequalities $V \ge \cos B_+$, $W \ge \cos C_+$, we obtain

$${\rm LHS}^2 - {\rm RHS}^2 \ge u^2+v^2+w^2 + 2uv\cos A_+ + 2vw \cos B_+ + 2uv\cos C_+\tag{*2}$$

Consider following $3$ vectors in $\mathbb{R}^2$,

$$\vec{u} = (u,0),\quad \vec{v} = (v\cos C_+,v\sin C_+),\quad \vec{w} = (w\cos B_+,-w\sin B_+)$$

It is easy to see $$\vec{u}\cdot\vec{v} = uv \cos C_+\quad\text{ and }\quad\vec{w}\cdot\vec{u} = wu \cos B_+$$ Using the fact $A_+ + B_+ + C_+ = 2\pi$, we find \begin{align}\vec{v}\cdot\vec{w} &= vw (\cos B_+\cos C_+ - \sin B_+\sin C_+)\\ &= vw\cos(B_+ + C_+) = vw\cos(2\pi - A_+) = vw\cos A_+\end{align}

Substitute these back into $(*2)$, we obtain

$${\rm LHS}^2-{\rm RHS}^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2\vec{u}\cdot\vec{v} + 2\vec{v}\cdot\vec{w} + 2\vec{w}\cdot\vec{u} = |\vec{u} + \vec{v} + \vec{w}|^2 \ge 0$$ From this, inequality $(*1)$ follows.

When $a_1 = a_2 = a, b_1 = b_2 = b, c_1 = c_2 = c$, we have $\Delta_1 = \Delta_2 = [ABC]$.
Inequality $(*1)$ reduces to the desired inequality: $$\bbox[padding: 1em;border:1px solid blue]{ xa^2 + yb^2 + zc^2 \ge 4 \sqrt{xy+yz+zx} [ABC]}$$

Notes

• $\color{blue}{[1]}$ - proof adapted from a chinese book 不等式探秘 (Questing for the Secrets of inequalities) by 李世杰, 李盛 (ISBN 978-7-5603-6228-1).