solution to a root inequality I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$,  $y=b^2+2ac$,  $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.
 A: I will assume that $a, b, c \geq 0$. Change to cylindrical coordinates with axis in the direction of $(1, 1, 1)^T$ as follows:
$$\left(\begin{matrix} a \\ b \\ c \end{matrix}\right) = \frac{z}{\sqrt{3}}\left(\begin{matrix} 1 \\ 1 \\ 1\end{matrix}\right) + \rho\left[\frac{\cos\phi}{\sqrt{6}}\left(\begin{matrix} 1 \\ 1 \\ -2\end{matrix}\right) + \frac{\sin\phi}{\sqrt{2}}\left(\begin{matrix} 1 \\ -1 \\ 0\end{matrix}\right)\right]$$
The inequality then becomes:
$$\sum_{j=1}^3 \sqrt{z^2+\rho^2\cos(2\phi+\frac{2\pi}{3}j)} \leq \sqrt{z^2+\rho^2}+2\sqrt{z^2-\frac{1}{2}\rho^2}$$
The left side depends on $\phi$, while the right side does not, and we have equality when $\phi = \frac{k\pi}{3}$ with $k$ an integer. Evaluating the partial derivative with respect to $\phi$ of the left side shows the existence of critical points at $\phi = \frac{k\pi}{6}$ for integer $k$. Plugging these in to the inequality shows that equality holds when $\phi = \frac{k\pi}{3}$, while strict inequality holds when $\phi = \frac{(2k+1)\pi}{6}$ (unless $\rho = 0$).
We first examine the case when $z^2 \geq \rho^2$. Consider the function
$$f(x_1, x_2, x_3) = \sum_{i=1}^3\sqrt{z^2 + \rho^2x_i}$$
subject to the constraints
$$g(x_1,x_2,x_3)=x_1 + x_2 + x_3 = 0$$
$$h(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2=\frac{3}{2}$$
The method of Lagrange multipliers dictates that local maxima will only be found where $\nabla f + \lambda \nabla g + \mu \nabla h = 0$. Then for each $x_i$ we have
$$\rho^2(z^2+\rho^2x_i)^{-1/2}+\lambda+2\mu x_i=0$$
Let $\psi(x) = \rho^2(z^2+\rho^2x)^{-1/2}+\lambda+2\mu x$. Since $\frac{d^2 \psi}{d x^2} = \frac{3}{4}\rho^6(z^2+\rho^2x)^{-5/2} > 0$, $\psi$ can have at most two roots, so local maxima can only be found when two of the $x_i$ are equal. This corresponds to $\phi = \frac{k\pi}{6}$ for integer $k$. Since we have equality at $\phi = \frac{k\pi}{3}$ and strict inequality at $\phi = \frac{(2k+1)\pi}{6}$, the critical points at $\phi=\frac{k\pi}{3}$ are maxima and the critical points at $\phi = \frac{(2k+1)\pi}{6}$ are local minima. Since the inequality holds at the maxima, it holds for all $\phi$.
Now consider the case when $z^2 < \rho^2$. We know that the term $z^2 - \frac{1}{2}\rho^2$ in the right-hand side is always well-defined, so $\frac{1}{2}\rho^2 \leq z^2 < \rho^2$. The left-hand side is then undefined within a neighborhood of $\phi = \frac{(2k+1)\pi}{6}$. Since $\phi = \frac{k\pi}{3}$ are maxima for $z^2 \geq \rho^2$, by continuity with respect to $z$ and the fact that critical points can only occur at $\frac{k\pi}{6}$, it follows that $\phi = \frac{k\pi}{3}$ are maxima for all $z^2 \geq \frac{1}{2}\rho^2$. Then since the inequality holds at the maxima, it holds for all $\phi$ such that it is well-defined, and thus for all $a, b, c \geq 0$. $\square$
A: We need to prove that
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc}\geq\sum\limits_{cyc}\sqrt{a^2+2bc}$$ or
$$\sum\limits_{cyc}\left(\sqrt{a^2+b^2+c^2}-\sqrt{c^2+2ab}\right)\geq2\left(\sqrt{a^2+b^2+c^2}-\sqrt{ab+ac+bc}\right)$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}}\geq\sum\limits_{cyc}\frac{(a-b)^2}{\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}}$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2\left(\sqrt{ab+ac+bc}-\sqrt{c^2+2ab}\right)}{\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}}\geq0$$ or
$$\sum\limits_{cyc}\frac{-(a-b)^2(c-a)(c-b)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\left(\tfrac{-(a-b)^2(c-a)(c-b)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}+\tfrac{(a-b)^2(c-a)(c-b)}{2\sqrt{ab+ac+bc}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}\right)}\right)\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)-2\sqrt{ab+ac+bc}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}\right)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(-\sqrt{(a^2+b^2+c^2)(ab+ac+bc)}+\sqrt{c^2+2ab}\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)+c^2+2ab-2(ab+ac+bc)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(\left(\sqrt{c^2+2ab}-\sqrt{ab+ac+bc}\right)\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)+c^2+2ab-(ab+ac+bc)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(c-a)(c-b)\left(\frac{(c-a)(c-b)\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)}{\sqrt{c^2+2ab}+\sqrt{ab+ac+bc}}+(c-a)(c-b)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(c-a)^2(c-b)^2\left(\frac{\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}}{\sqrt{c^2+2ab}+\sqrt{ab+ac+bc}}+1\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$
Done!
