Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$

Question

Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$

I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I just thought of using $uvw$, but I am not allowed to use it during my exam. Because of the equality I cannot use AMGM, Cauchy-Schwarz, etc.

I tried to use Mixing-Variables, but I failed. Please help.

Answer 1

For $x=y=\frac{1}{3}$ and $z=\frac{5}{3}$ we obtain a value $\frac{25}{27}.$

We'll prove that it's a minimal value.

Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, the condition gives $$3u^2-2v^2=1$$ and does not depend on $w^3$, which says that we need to find a minimum of the linear function of $w^3$, which happens in our case for the maximal value of $w^3$.

Now, $x$, $y$ and $z$ they are roots of the equation: $$(t-x)(t-y)(t-z)=0$$ or $$t^3-3ut^2+3v^2t=w^3,$$ which says that a graph of $f(t)=t^3-3ut^2+3v^2t$ and a graph of $g(t)=w^3$ have three common points.

Now, $$f'(t)=3t^2-6ut+3v^2=3(t^2-2ut+v^2),$$ which gives $$t_{max}=u-\sqrt{u^2-v^2}>0$$ and $$t_{min}=u+\sqrt{u^2-v^2}>0.$$ Also, we see that $f(0)=0$.

Now, we can draw a graph of $f$ and we see that $w^3$ gets a maximal value, when $g$ is a tangent line to graph of $f$, which happens for equality case of two variables.

Let $y=x$.

Thus, $z=\sqrt{3-2x^2},$ where $0<x<\sqrt{1.5}$ and we need to prove that $$(2-x)^2(2-\sqrt{3-2x^2})\geq\frac{25}{27}$$ or $$\frac{191}{27}-8x+2x^2\geq(2-x)^2\sqrt{3-2x^2}$$ or $$(3x-1)^2(162x^4-1188x^3+3159x^2-3594x+1489)\geq0$$ and it's enough to prove that $$162x^4-1188x^3+3159x^2-3594x+1489\geq0,$$ which is true because $$162x^4-1188x^3+3159x^2-3594x+1489\geq$$ $$\geq162x^4-1188x^3+3158x^2-3594x+1488=$$ $$=2(81x^4-594x^3+1579x^2-1797x+744)=$$ $$=2\left((9x^2-33x+27)^2+4x^2-15x+15\right)>0.$$ Now we see that $\frac{25}{27}$ is a minimal value of the expression $\prod\limits_{cyc}(2-x)$ for any reals $x$, $y$ and $z$ such that $x^2+y^2+z^2=3.$

Answer 2

From Inequality on AoPS:

Without loss of generality $x \ge y \ge z > 0$, so that $z \le 1$. We have $$ 2(2-x)(2-y) = (x+y-2)^2 +4 - x^2-y^2 \ge 4 - x^2 - y^2 = 1+z^2 $$ and therefore $$ (2-x)(2-y)(2-z) \ge \frac 1 2 (1+z^2)(2-z) =: f(z) \, . $$ An elementary calculation shows that the minimum of $f$ on $[0, 1]$ is $f(1/3) = 25/27$, i.e. $$ (2-x)(2-y)(2-z) \ge \frac{25}{27} \, . $$ Equality holds if $(x, y, z)$ is a permutation of $(5/3, 1/3, 1/3)$.

Answer 3

Wlog $z\le 1$ because the variables cannot all be $\ge1$. For fixed $z$, we want to minimize $(2-x)(2-y)$ under a constraint $x^2+y^2=3-z^2$, which is a constant between $2$ and $3$.

Now $$ \begin{align}(2-x)(2-y)&=4-2(x+y)+xy\\&=4-2(x+y)+\frac12(x+y)^2-\frac{3-z^2}2\\ &=\frac12\left(x+y-2\right)^2+2-\frac{3-z^2}2 \\ &\ge2-\frac{3-z^2}2\\&=\frac{z^2+1}2 \end{align}$$ with equality iff $x+y=2$. Note that this makes $x^2+y^2=x^2+(2-x)^2=4-4x+2x^2=2(x-1)^2+2$, i.e., we can always find such $x,y$ when $z\le 1$ and $x^2+y^2=3-z^2\in [2,3]$. So we want to minimize $\frac{z^2+1}2\cdot(2-z)=\frac12(2-z+2z^2-z^3)$ with $0<z\le1$. The derivative of this cubic is $-3z^2+4z-1=(z-1)(1-3z)$, so we find the desired minimum at $z=\frac13$.

Answer 4

There is also a straight-forward solution with Lagrange multipliers. It is more convenient to examine the (continuous) function $$ f(x, y, z) = (2-x)(2-y)(2-z) $$ on the (compact) set $K = \{ (x, y, z) \in \Bbb R^3 \mid x^2+y^2+z^2 = 3 \}$ first, without the restriction to positive arguments.

$f$ attains both minimum and maximum on $K$, and the Lagrange method gives that at the extremal points $$ -(2-y)(2-z) = \lambda 2x \\ -(2-x)(2-z) = \lambda 2y \\ -(2-x)(2-y) = \lambda 2z $$ for some $\lambda \in \Bbb R$. We see that $\lambda$ cannot be zero, so that $$ x(2-x) = y(2-y) = z(2-z) =: \alpha \, , $$ i.e. $x, y, z$ are all solutions of the quadratic equation $t^2 - 2t + \alpha = 0$. It follows that any two of them are equal or add to $2$. So we have two possible cases:

1. $x=y=z$. Then necessarily $x=y=z = \pm 1$.

2. Up to a permutation, $x=y=2-z$. Substituting that into $x^2+y^2+z^2$ gives $$ 3 = 2x^2 + (2-x)^2 = 3x^2 -4x + 4 \iff (3x-1)(x-1) = 0 \, . $$ If $x=1$ then (again) $x=y=z=\pm 1$. Otherwise $(x, y, z) = (1/3, 1/3, 5/3)$.

So the only possible extremal values of $f$ on $K$ are $$ \begin{align} f(1, 1, 1) &= 1 \, \\ f(-1, -1, -1) &= 27 \, \\ f(1/3, 1/3, 5/3) &= 25/27 \, , \end{align} $$ and we conclude that $25/27$ is the minimum.

This is also the mimimum for the given problem because it is attained at a point with $x, y, z > 0$.