Interesting Inequality 2

The expression can be rewritten as $2\big[(x-y)^2 + (y-z)^2 + (z-w)^2 + (w-x)^2\big] + (xy-5)^2 + (yz-5)^2 + (zw-5)^2 + (wx-5)^2 + 96$ and hence the minimum value is $\boxed{96}$ , achieved when $x=y=z=w=\sqrt{5}$ .

$\begin{aligned} X & = \blue{4(x^2+y^2+z^2+w^2)} + \red{(xy-7)^2 + (yz-7)^2 + (zw-7)^2 + (wx-7)^2} & \small \blue{\text{By AM-GM inequality}} \\ & \ge \blue{16\sqrt[4]{x^2y^2z^2w^2}} + \red{\frac {(xy-7+yz-7+zw-7+wx-7)^2}4} & \small \red{\text{By Titu's lemma}} \\ & \ge 16\sqrt{xyzw} + \frac {(\red{xy+yz+zw+wx}-28)^2}4 & \small \red{\text{By AM-GM inequality}} \\ & \ge 16\sqrt{xyzw} + \frac {(\red{4\sqrt{xyzw}}-28)^2}4 & \small \blue{\text{Equality when }x=y=z=w=\sqrt 5} \\ & = 16 \times 5 + \frac {(4\times 5-28)^2}4 \\ & = \boxed{96} \end{aligned}$

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References:

• AM-GM inequality
• Titu's lemma

$f(x,y,z,w)=4(x^2+y^2+z^2+w^2)+(xy-7)^2+(xz-7)^2+(zw-7)^2+(xw-7)^2$ Solve system

$f_x=x(8+2y^2+2w^2)-14(w+y)=0$

$f_y=y(8+2x^2+2z^2)-14(x+z)=0$

$f_z=z(8+2y^2+2w^2)-14(w+y)=0$

$f_w=w(8+2x^2+2z^2)-14(x+z)=0$

Find

$x=\frac {7(w+y)}{4+y^2+w^2}$

$y=\frac {7(x+z)}{4+x^2+z^2}$

$z=\frac {7(w+y)}{4+y^2+w^2}$

$w=\frac {7(x+z)}{4+x^2+z^2}$

From here we see $x=z$ and $y=w$ and $min f(x,y,z,w)=min f(x,y,x,y)= min g(x,y)$

$g(x,y)=8(x^2+y^2)+4(xy-7)^2$

Solve system

$g_x=16x+8(xy-7)y=0$

$g_y=16y+8(xy-7)x=0$

we find

1. $x=-\sqrt{5}, y=-\sqrt{5}$

2. $x=\sqrt{5}, y=\sqrt{5}$

3. $x=0,y=0$

And find min $f(x,y,z,w)=96$ .

Control

$d^2g(x,y)=Ag_{xx} {dx}^2+Bg_{xy}{dx dy}+Cg_{yy}{dy}^2 \geq 0$ for $x=\sqrt{5}, y=\sqrt{5}$ .

R. M. S. - A. M. inequality :

$4(x^2+y^2+z^2+w^2)\geq (x+y+z+w)^2$

$(xy-7)^2+(yz-7)^2+(zw-7)^2+(wx-7)^2\geq \dfrac{\left ((xy-7)+(yz-7)+(zw-7)+(wx-7)\right )^2 }{4}=\dfrac{(xy+yz+zw+wx-28)^2}{4}$

A. M. - G. M. inequality :

$(x+y+z+w)^2\geq 16\sqrt {xyzw}$

$xy+yz+zw+wx\geq 4\sqrt {xyzw}$

So, $4(x^2+y^2+z^2+w^2)+(xy-7)^2+(yz-7)^2+(zw-7)^2+(wx-7)^2\geq 16\sqrt {xyzw}+4(\sqrt {xyzw}-7)^2= 4(\sqrt {xyzw}-5)^2+96\geq 96$

The equality holds when $x=y=z=w$

Hence the required minimum of the given expression is $\boxed {96}$ .

I'd love to see a problem like this one where the solution isn't found at some point where all the variables have the same value, but that doesn't appear to be the case here.

If we look at the line where $w=x=y=z$ we are basically trying to minimize the function $f(x) = 16x^2 + 4(x^2 - 7)^2$ , which has a minimum at $x=\sqrt{5}$ , where its value is $96$ . I did plug the 4-variable function into an Excel worksheet, where I tested the neighborhood of this point and satisfied myself that at least this was a local minimum.

I'd love to see an actual proof.