solution to a root inequality

Question

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.

Answer 1

$x=a-b,y=b-c,z=c-a,u^2=a^2+b^2+c^2,p^2=ab+bc+ac \to x+y+z=0,x^2+y^2+z^2=2u^2-2p^2,xy+yz+xz=p^2-u^2 \to x^2y^2+y^2z^2+x^2z^2=(p^2-u^2)^2,(xy)^4+(yz)^4+(xz)^4=(u^2-p^2)^2+(4p^2-4u^2)x^2y^2z^2,u^2-x^2-p^2=-yz,u^2-y^2-p^2=-xz,u^2-z^2-p^2=-xy$

$\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} \iff $

$u+2p \ge \sqrt{u^2-x^2}+\sqrt{u^2-y^2}+\sqrt{u^2-z^2} \iff$

$ 2up \ge \sqrt{u^2-x^2}\sqrt{u^2-y^2}+\sqrt{u^2-x^2}\sqrt{u^2-z^2}+\sqrt{u^2-z^2}\sqrt{u^2-y^2}-p^2 \iff $

$ 4u^2p^2 \ge \sum_{cyc}(u^2-x^2)(u^2-y^2)+p^4+2\sum_{cyc}(u^2-x^2)\sqrt{u^2-z^2}\sqrt{u^2-y^2}-2p^2\sum_{cyc}\sqrt{u^2-x^2}\sqrt{u^2-y^2} \iff $

$ 4u^2p^2 \ge 3u^4+p^4+\sum_{cyc}x^2y^2-2(x^2+y^2+z^2)u^2+2\sum_{cyc}(u^2-x^2-p^2)\sqrt{u^2-z^2}\sqrt{u^2-y^2} \iff p^2(u^2-p^2) \ge -\sum_{cyc}yz\sqrt{u^2-z^2}\sqrt{u^2-y^2} \iff p^2(u^2-p^2)+xy\sqrt{u^2-x^2}\sqrt{u^2-y^2} \ge -yz\sqrt{u^2-z^2}\sqrt{u^2-y^2}-xz\sqrt{u^2-x^2}\sqrt{u^2-z^2} \iff [p^2(u^2-p^2)]^2+2x^2y^2(u^2-x^2)(u^2-y^2)-\sum_{cyc}x^2y^2(u^2-x^2)(u^2-y^2) \ge 2xy\sqrt{u^2-x^2}\sqrt{u^2-y^2}[z^2(u^2-z^2)-p^2(u^2-p^2)] \iff x^2y^2(2p^4+2z^4-u^2z^2)\ge 2x^2y^2\sqrt{u^2-x^2}\sqrt{u^2-y^2}(p^2-z^2) \iff x^4y^4[((2p^4+2z^4-u^2z^2)^2-4(p^2-z^2)^2(p^4+z^4-u^2z^2+2p^2z^2)]\ge 0 \iff (xyz)^4(4p^2-u^2)^2\ge 0 $when and only when $x=0 $ or $ y=0 $ or $z=0 $ it takes $"="$ QED.

EDIT: I add some notes for the process:

1. $xy>0 \to -yz\sqrt{u^2-z^2}\sqrt{u^2-y^2}-xz\sqrt{u^2-x^2}\sqrt{u^2-z^2}>0$
2. if $p^2 < z^2$, then the inequality already OK as $2p^4+2z^4-u^2z^2 \ge p^4+z^4-u^2z^2+2p^2z^2=(u^2-x^2)(u^2-y^2) \ge 0$

Answer 2

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!

Answer 3

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 $a, b, c \geq 0$. $\square$