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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: $$a^2x+b^2y+c^2z\geq 4[ABC]\sqrt{xy+xz+yz}$$ $$a^2x+b^2y+c^2z\geq \frac{abc}{R}\sqrt{xy+xz+yz}$$ $$\frac{aRx}{bc}+\frac{bRy}{ac}+\frac{cRz}{ab}\geq \sqrt{xy+xz+yz}$$ $$\frac{1}{2}\left(\frac{4aR^2x}{2Rbc}+\frac{4bR^2y}{2Rac}+\frac{4cR^2z}{2Rab}\right)\geq \sqrt{xy+xz+yz}$$ $$x\frac{\sin\alpha }{\sin\beta \sin \gamma}+y\frac{\sin\beta }{\sin\alpha \sin \gamma}+z\frac{\sin\gamma }{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}$$

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

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

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

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

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

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:

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

\begin{equation} \cot\alpha \geq \cot \alpha' \tag{3} \end{equation} \begin{equation} \cot\beta \geq \cot \beta' \tag{4} \end{equation} \begin{equation} \cot\gamma'\geq \cot \gamma \tag{5} \end{equation} 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*} \begin{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} \end{equation}

\begin{equation*} f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')= \end{equation*} \begin{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} \end{equation}

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

\begin{equation} 0 \geq \cot\alpha'-\cot\alpha \tag{8} \end{equation}

\begin{equation} 0 \geq \cot \beta' -\cot\beta \tag{9} \end{equation}

\begin{equation} \cot\gamma'-\cot\gamma \geq 0 \tag{10} \end{equation} 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:

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

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:

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

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 $

\begin{equation} \frac{a'R'}{b'c'}\geq \frac{aR}{bc} \tag{13} \end{equation}

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

Suppose by contradiction that it occurs:

\begin{equation} \cot\alpha+\cot\gamma> \cot\alpha'+\cot\gamma' \tag{15} \end{equation}

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:

\begin{equation} \cot\beta+\cot\gamma> \cot\beta'+\cot\gamma' \tag{16} \end{equation}

See that:

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

This contradicts the inequality (13).Therefore:

\begin{equation} \cot\alpha+\cot\gamma \leq \cot\alpha'+\cot\gamma' \tag{17} \end{equation}

\begin{equation} \cot\beta+\cot\gamma \leq \cot\beta'+\cot\gamma' \tag{18} \end{equation}

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:

\begin{equation} (\cot\alpha+\cot\gamma) (\cot\beta'-\cot\beta)\geq (\cot\alpha'+\cot\gamma')(\cot\beta'-\cot\beta) \tag{19} \end{equation}

\begin{equation} (\cot\beta+\cot\gamma) (\cot\alpha'-\cot\alpha)\geq (\cot\beta'+\cot\gamma')(\cot\alpha'-\cot\alpha) \tag{20} \end{equation}

On the other hand of inequalities (3) and (4) we know that: \begin{equation} \cot\alpha+\cot\beta \geq \cot\alpha'+\cot\beta' \tag{21} \end{equation} 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:

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

\begin{equation*} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq \end{equation*} \begin{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} \end{equation}

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:

\begin{equation} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')+f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq 0 \end{equation} And this implies, finally, that:

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

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  • $\begingroup$ 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. $\endgroup$ Dec 26, 2016 at 3:05
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    $\begingroup$ A proof that does not use algebraic inequalities... $\endgroup$ Dec 26, 2016 at 3:09
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    $\begingroup$ What a work since your recent post ! $\endgroup$
    – Jean Marie
    Dec 26, 2016 at 3:44
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    $\begingroup$ @Israel Meireles Chrisostomo There is a smooth and an easy algebraic proof. $\endgroup$ Dec 26, 2016 at 7:17
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    $\begingroup$ @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. $\endgroup$ Dec 26, 2016 at 7:26

3 Answers 3

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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!

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

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  • $\begingroup$ 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. $\endgroup$
    – dezdichado
    Aug 31, 2022 at 2:02
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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).
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