Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$

Question

Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$. By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.

Question from Jalil Hajimir

Answer 1

Let $x=\frac{a}{\sqrt2},$ $y=\frac{b}{\sqrt2}$ and $z=\frac{c}{\sqrt2}.$

Thus, since we can assume that $(a^2-1)(b^2-1)\geq0,$ by C-S we obtain: $$\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y+z)^2}=\frac{(a^2+2)(b^2+2)(c^2+2)}{4(a+b+c)^2}\geq$$ $$\geq\frac{3(a^2+b^2+1)(1+1+c^2)}{4(a+b+c)^2}\geq\frac{3(a+b+c)^2}{4(a+b+c)^2}=\frac{3}{4}.$$ The equality occurs for $x=y=z=\frac{1}{\sqrt2},$ which says that we got a minimal value.

Answer 2

You first fix $y, z$ and let $x > 0$ vary. Taking derivative with respect to $x$, dropping all those nonnegative terms such as $y^2+1$ to simplify notation, leads to $$ \frac{d (OP\ full\ epxr)}{d x} \approx x - \frac{(x^2+1)}{x+y+z} = \frac{x(y+z) - 1}{x+y+z}, $$ where the $\approx$ means I dropped some positive terms (they do not affect my analysis of the positivity of the derivate).

It is evident that the gradient is negative for small $x$, and once $x > \frac{1}{y+z}$ the gradient becomes positive. Hence the function is minimized at $x = \frac{1}{y+z}$ when $y, z$ are fixed. Similarly, the function is minized at $y = \frac{1}{x+z}$ when $x, z$ are fixed. And the function is minimized at $z = \frac{1}{x+y}$ when $x, y$ are fixed.

Let the global minimizing point be $x_0, y_0, z_0$, given the previous arguments, we must have $x_0= \frac{1}{y_0+z_0}, y_0 = \frac{1}{x_0+z_0}, z_0 = \frac{1}{x_0+y_0} \Rightarrow x_0 = y_0 = z_0 = \frac{\sqrt{2}}{2}$ (otherwise, we can find a point with smaller value).

Hence the global minimum is unique at $x = y = z = \frac{\sqrt{2}}{2}$ if one exists.

For rigorous argument that the a global minimum exists, one can look at a compact set $[\epsilon, N]\times[\epsilon, N]\times [\epsilon, N]$. The function must have a global minimum in the compact set. One can easily argue it does not take minimum at the boundary (contradicts requirements previously stated, or compare function value on the boundary with that of $x = y = z = \frac{\sqrt{2}}{2}$). Hence the minimum MUST be in the interior (with gradients being zero).

Hence $x = y = z = \frac{\sqrt{2}}{2}$ is the unique global minimum.

Answer 3

If you want some calculus/analysis argument:

After establishing there must exist a global minimum, let $p$ be the global minimum. Then we must have that $$f(x) = x^2\left((y^2+1)(z^2+1) - p\right) - 2xp(y+z) + (y^2+1)(z^2+1) - p(y+z)^2\geq 0$$ as a quadratic in $x.$ So the discriminant is non-positive: $$D =4\left[p^2(y+z)^2 - (y^2+1)^2(z^2+1)^2 - p^2(y+z)^2+(y^2+1)(z^2+1)p(1+(y+z)^2)\right]\leq 0\iff $$ $$p\leq\min\dfrac{(y^2+1)(z^2+1)}{1+(y+z)^2}.$$ But $$4(y^2+1)(z^2+1) - 3 - 3(y+z)^2 = 4y^2z^2+y^2+z^2-6yz+1 = (y-z)^2+(2yz-1)^2\geq 0.$$ So $p = \dfrac{3}{4}$ by continuity argument and it is achieved by $y = z = \dfrac{1}{\sqrt{2}},$ which in return easily tells us that that $x$ is also $\dfrac{1}{\sqrt{2}}$ for the minimum to be attained.