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

