Let $x,y,z > 0$ and $x^2+y^2+z^2=3$, prove that $$(x+2y+z)\cdot \left( \frac{x}{y} +\frac{2y}{z}+\frac{z}{x}\right) > 12.$$
The coefficient $2$ destroys the symmetry of this inequality and makes the use of C-S to be impossible. I guess someone experiences with classical inequality can help. Thanks
Remark: Here is a proof without using calculus. I use AM-GM basically. So I guess there is better proof using AM-GM only.
After homogenization, it suffices to prove that $$(x + 2y + z)^2\left( \frac{x}{y} +\frac{2y}{z}+\frac{z}{x}\right)^2 > 48x^2 + 48y^2 + 48z^2. \tag{1} $$
Using $(a+b+c)^2 \ge 3(ab + bc + ca)$, we have $$\left( \frac{x}{y} +\frac{2y}{z}+\frac{z}{x}\right)^2 \ge 3\left(\frac{2x}{z} +\frac{2y}{x}+\frac{z}{y}\right). \tag{2}$$
From (1) and (2), it suffices to prove that $$(x + 2y + z)^2\left(\frac{2x}{z} +\frac{2y}{x}+\frac{z}{y}\right) > 16x^2 + 16y^2 + 16z^2. \tag{3} $$
Since (3) is homogeneous, assume that $z = 1$. It suffices to prove that $$(x + 2y + 1)^2\left( 2x +\frac{2y}{x}+\frac{1}{y}\right) > 16x^2 + 16y^2 + 16. \tag{4} $$
We have \begin{align*} &(x + 2y + 1)^2\left( 2x +\frac{2y}{x}+\frac{1}{y}\right)\\ ={}& 2\,{x}^{3}+8\,{x}^{2}y+8\,x{y}^{2}+4\,{x}^{2}+10\,xy+8\,{y}^{2}+6\,x+8y + 4\\[6pt] &\quad + {\frac {{8y}^{3}}{x}}+{\frac {{8y}^{2}}{x}}+{\frac {2y}{x}}+{ \frac {{x}^{2}}{y}}+{\frac {2x}{y}}+\frac{1}{y}\\[6pt] ={}& \left(8xy + \frac{8y^3}{x} \right) + (2x^3 + 6x) + 4x^2 + \left( 8x^2y + \frac{x^2}{y}\right) \\[6pt] &\quad + \left[\left(2y + \frac{2}{y} + 8y^2\right)x + \frac{8y^2 + 2y}{x}\right] + 8y^2 + 8y + \frac{1}{y} + 4 \\[6pt] \ge{}& 16y^2 + 2\sqrt{12}\, x^2 + 4x^2 + 2\sqrt{8}\, x^2 + 2\sqrt{ \left(2y + \frac{2}{y} + 8y^2\right)\cdot(8y^2 + 2y) }\\ &\quad + 8y^2 + 8y + \frac{1}{y} + 4\\[6pt] \ge{}& 16y^2 + 16x^2 + 2\sqrt{ \left(2y + \frac{2}{y} + 8y^2\right)\cdot(8y^2 + 2y) } + 8y^2 + 8y + \frac{1}{y} + 4 \tag{5} \end{align*} where we use AM-GM to obtain $8xy + \frac{8y^3}{x} \ge 16y^2$, and $2x^3 + 6x \ge 2\sqrt{12}\, x^2$, and $8x^2y + \frac{x^2}{y} \ge 2\sqrt{8}\, x^2$ etc., and we use $2\sqrt{12} + 4 + 2\sqrt{8} > 16$.
From (4) and (5), it suffices to prove that $$2\sqrt{ \left(2y + \frac{2}{y} + 8y^2\right)\cdot(8y^2 + 2y) } + 8y^2 + 8y + \frac{1}{y} - 12 > 0 \tag{6}$$ which is true. The proof of (6) is given at the end.
We are done.
Proof of (6):
(6) is written as $$4\sqrt{ \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y) } + 8y^2 + 8y + \frac{1}{y} - 12 > 0.$$
Using $\sqrt{u} = \frac{2u}{2\sqrt u} \ge \frac{2u}{1 + u}$ for all $u > 0$ (by AM-GM), we have $$\sqrt{ \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y) } \ge \frac{2 \cdot \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y) }{1 + \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y)}.$$
It suffices to prove that $$4 \cdot \frac{2 \cdot \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y) }{1 + \left(y + \frac{1}{y} + 4y^2\right)\cdot(4y^2 + y)} + 8y^2 + 8y + \frac{1}{y} - 12 > 0$$ or $${\frac {128\,{y}^{7}+192\,{y}^{6}+8\,{y}^{5}+24\,{y}^{4}+52\,{y}^{3}+{ y}^{2}-12\,y+2}{16\,{y}^{5}+8\,{y}^{4}+{y}^{3}+4\,{y}^{2}+2\,y}} > 0$$ which is true.
We are done.
Let $$x=u^2,\quad y=v^2,\quad z=w^2,\qquad(1)$$ then the inequality becomes $$(u^2+2v^2+w^2)\left(\dfrac{u^2}{v^2}+\dfrac{2v^2}{w^2}+\dfrac{w^2}{u^2}\right) > 12$$ for $$u^4+v^4+w^4=3.$$ Using Cauchy-Schwarz inequality, we have $$(u^2+2v^2+w^2)\left(\dfrac{u^2}{v^2}+\dfrac{2v^2}{w^2}+\dfrac{w^2}{u^2}\right)\geq\left(\dfrac{u^2}v+\dfrac{2v^2}w+\dfrac{w^2}u\right)^2.\qquad (2)$$ Therefore, it is sufficiently to find the conditional minimum of $$\dfrac{u^2}v+\dfrac{2v^2}w+\dfrac{w^2}u$$ for $u,v,w>0$ and $u^4+v^4+w^4=3.$
Apply the method of Lagrange multipliers, which reduces it to a system of equations.
For this, search uncoditional minimum of function
$$f(u,v,w,\lambda) = \dfrac{u^2}v+\dfrac{2v^2}w+\dfrac{w^2}u + \lambda\left(u^4+v^4+w^4-3\right).$$
The nesessary conditions of extremum are
$$f'_u=0,\quad f'_v=0,\quad f'_w=0,\quad f'_\lambda=0,$$
so
$$\begin{cases}
\dfrac{2u}v-\dfrac{w^2}{u^2} + 4\lambda u^3 = 0\\
\dfrac{4v}w-\dfrac{u^2}{v^2} + 4\lambda v^3 = 0\\
\dfrac{2w}u-\dfrac{2v^2}{w^2} + 4\lambda w^3 = 0\\
u^4+v^4+w^4-3 = 0.
\end{cases}$$
Given $u,v,w>0,$
$$\begin{cases}
\dfrac{2u^2}v-\dfrac{w^2}u + 4\lambda u^4 = 0\\
\dfrac{4v^2}w-\dfrac{u^2}v + 4\lambda v^4 = 0\\
\dfrac{2w^2}u-\dfrac{2v^2}w + 4\lambda w^4 = 0\\
u^4+v^4+w^4-3 = 0.
\end{cases}$$
Now we can eliminate $\lambda:$
$$\begin{cases}
\left(\dfrac{2u^2}v-\dfrac{w^2}u\right)v^4=\left(\dfrac{4v^2}w-\dfrac{u^2}v\right)u^4\\
\left(\dfrac{2u^2}v-\dfrac{w^2}u\right)w^4=\left(\dfrac{2w^2}u-\dfrac{2v^2}w\right)u^4\\
u^4+v^4+w^4-3 = 0.
\end{cases}$$
For $u,v,w>0$
$$\begin{cases}
2u^3v^4w-v^5w^3=4u^5v^3-u^7w\\
2u^3w^5-vw^7=2u^4vw^3-2u^5v^3\\
u^4+v^4+w^4-3 = 0.
\end{cases}$$
$$\begin{cases}
u^3w(u^4+2v^4) = 4v^3u^5+v^5 w^3\\
2u^3(u^2v^3+w^5) = vw^3(2u^4+w^4)
u^4+v^4+w^4-3 = 0.
\end{cases}$$
The first two equations are homogenius, and we can use substituion $$s=\dfrac vu,\quad t=\dfrac wu,$$ $$\begin{cases} s^5t^3 - (2s^4+1)t + 4s^3 = 0 \\ st^7 - 2t^5 + 2st^3 - 2s^3 = 0. \end{cases}\qquad (3)$$ This system has the only extremum in the area $$s\approx 0.739672,\quad t\approx 1.36409.$$ Using the third equation, we obtain the extremum $$(u_0,v_0,w_0,f_0)=(0.890923, 0.65899, 1.215299, 3.576929).$$
Note that in the edges of area limit of the target function is $+\infty$ and the only extremum in the area, this is global minimum in area, wherein $f_0^2\approx 12.794424.$
Using $(1-2)$ and taking in account the computation accuracy this means that $$\boxed{(x+2y+z)\left(\dfrac{x}y+\dfrac{2y}z+\dfrac{z}x\right) > 12}$$
Addition
Considering the unknown $s$ as a parameter, it is possible to reduce the order of the system $(3)$ with respect to unknown $t$. Finally, we can obtain an equation of one unknown $s$.
In fact, instead of the second equation of the system can take a linear combination of the first and second equations. So, it is possible to multiply the second equation of system (3) on the leading coefficient of the first equation and subtract from it the first equation, multiplied to the leading coefficient of the second equation, which leads to a decrease in the order of the second equation. By repeating this action if necessary, the second equation may be obtained with less unknown $t$ than the first.
For $s\not=0,$ obtaining $$\begin{cases} s^5t^3 - (2s^4+1)t + 4s^3 = 0 \\ C_{2,2}t^2 + C_{2,1}t +C_{2,0}= 0,\\ \end{cases}$$ where $$C_{2,2} = 8s^{12}+8s^8,$$ $$C_{2,1} = -4s^{14}-16s^{11}-2s^{10}-4s^8-4s^4-1,$$ $$C_{0,0} = 2s^{17}+8s^{13}+8s^7+4s^3.$$
Similarly, there is posible to reduce the level of the first through the second equation with obtaining the system in the form $$\begin{cases} C_{1,1}t + C_{1,0} = 0\\ C_{2,2}t^2 + C_{2,1}t +C_{2,0}= 0,\\ \end{cases}$$ where $$C_{1,1} = -16s^{29} +16s^{28} +48s^{25} +16s^{24} -128s^{23} +288s^{22} +4s^{20} -256s^{19} +48s^{18} +16s^{16} -224s^{15} +24s^{14} +32s^{12} -64s^{11} +4s^{10} +24s^8 +8s^4 +1,$$ $$C_{1,0} = 256s^{22} -32s^{28} -36s^{27} -8s^{25} -128s^{24} -16s^{23} -8s^{31} -72s^{21} +384s^{18} -66s^{17} -32s^{15} +192s^{14} -16s^{13} -48s^{11} -24s^7-4s^3.$$
Again lowering the level of the second equation, we can obtain $$\begin{cases} C_{1,1}t + C_{1,0} = 0\\ C_{0,0} = 0,\\ \end{cases}$$ where $$C_{0,0} = 32s^{40} -56s^{43} -64s^{41} -4s^{47} -292s^{39} -144s^{37} +112s^{36} -944s^{35} +2112s^{34} -472s^{33} +152s^{32} -1856s^{31} +8192s^{30} -1472s^{29} +132s^{28} -10880s^{27} +14240s^{26}-2040s^{25} +176s^{24} -19616s^{23} +12384s^{22} -1216s^{21} +1300s^{20} -11056s^{19} +4260s^{18} -256s^{17} +2328s^{16} -1600s^{15} +40s^{14} +1194s^{12} -496s^{11} +4s^{10} +61s^8-64s^7+12s^4+1$$ Converting the system was carried out in vector form using the Mathcad program. Using $polyroots()$ function allows us to find all the roots on the Laguerre method, and among them is the only real root $$s_{38}\approx 0.73967226,$$ wherein $$t_{38} = \left.-\dfrac{C_{1,0}}{C_{1,1}}\right|_{s=s_{38}} \approx 1.36409053.$$
It is easily seen that the obtained solution satisfies $(3)$.
Another modification of the method used here.
