Inequality Proof

Solution Explanation

We wish to prove that for all real numbers $x, y, z$ satisfying

the following inequality holds:

Equivalently, we want to show

Key idea

Define the function

We want to prove that, subject to the constraint $x^2 + y^2 + z^2 = 2,$ we have

In other words, $2$ is an upper bound for $x + y + z - xyz$ on the surface of the sphere $x^2 + y^2 + z^2 = 2$.

A quick guess at the extremal points

A useful strategy with symmetric-like inequalities (or ones that impose a constraint such as $x^2 + y^2 + z^2 = 2$) is to look at “nice” or “extreme” points on the sphere:

1. Try $(1,1,0)$:

   • Then $x^2 + y^2 + z^2 = 1 + 1 + 0 = 2$, so the constraint is satisfied.
   • Compute $f(1,1,0)$: $$f(1,1,0) = 1 + 1 + 0 - (1 \cdot 1 \cdot 0) = 2.$$

   Thus at the point $(1,1,0)$, we achieve the value $2$.

2. Try $(\sqrt{2},0,0)$ (or any permutation):

   • Then $x^2 + y^2 + z^2 = 2 + 0 + 0 = 2$.
   • Compute $f(\sqrt{2},0,0)$: $$f(\sqrt{2},0,0) = \sqrt{2} + 0 + 0 \;-\; (\sqrt{2}\cdot 0\cdot 0) = \sqrt{2} < 2.$$

   This is clearly smaller than 2.

3. Try $\bigl(\sqrt{\tfrac{2}{3}},\sqrt{\tfrac{2}{3}},\sqrt{\tfrac{2}{3}}\bigr)$:

   • Then $x^2 + y^2 + z^2 = 3 \times \frac{2}{3} = 2$.
   • Compute $f$ there: $$x + y + z \;=\; 3 \sqrt{\tfrac{2}{3}} = \sqrt{6}, \quad xyz = \bigl(\sqrt{\tfrac{2}{3}}\bigr)^3 = \tfrac{2\sqrt{2}}{3\sqrt{3}}.$$ Numerically, $\sqrt{6} \approx 2.449$ and $\frac{2\sqrt{2}}{3\sqrt{3}} \approx 0.54.$ So $$f\bigl(\sqrt{\tfrac{2}{3}}, \sqrt{\tfrac{2}{3}}, \sqrt{\tfrac{2}{3}}\bigr) \;\approx\; 2.449 \;-\; 0.54 \;=\; 1.909,$$ which is still less than 2.

From these trials, it appears that the maximum value of $f$ on the sphere is $2$, and it is achieved for example at $(1,1,0)$ (or permutations thereof).

More rigorous justification (outline)

A standard rigorous approach uses the method of Lagrange multipliers to find the extrema of

subject to the constraint

1. Set up the gradient equations:
   We want $\nabla f = \lambda \,\nabla g$. Compute the gradients:

   $$\nabla f(x,y,z) = \Bigl(1 - yz,\; 1 - xz,\; 1 - xy\Bigr),$$ $$\nabla g(x,y,z) = \bigl(2x,\; 2y,\; 2z\bigr).$$

   The Lagrange multiplier condition is:

   $$\begin{cases} 1 - yz = 2 \lambda x,\\ 1 - xz = 2 \lambda y,\\ 1 - xy = 2 \lambda z,\\ x^2 + y^2 + z^2 = 2. \end{cases}$$

2. Check the point $(1,1,0)$:
   Plug in $(x,y,z)=(1,1,0)$:

   • Left-hand sides in the gradient equations: $$1 - (1 \cdot 0) = 1, \quad 1 - (1 \cdot 0) = 1, \quad 1 - (1 \cdot 1) = 0.$$
   • Right-hand sides become: $$2 \lambda \cdot 1 = 2\lambda, \quad 2 \lambda \cdot 1 = 2\lambda, \quad 2 \lambda \cdot 0 = 0.$$

   If we set $\lambda = \tfrac12$, these match perfectly:

   $$1 = 2\lambda = 2 \cdot \tfrac12 = 1, \quad 0 = 0.$$

   And clearly $1^2 + 1^2 + 0^2 = 2.$
   Thus $(1,1,0)$ (and any permutation of those coordinates) is indeed a critical point satisfying the constraint.

3. Conclude the maximum value:
   By checking various points and using the fact that $(1,1,0)$ is a critical point that yields $f(1,1,0) = 2$, one concludes that

   $$\max_{\,x^2 + y^2 + z^2 = 2} \bigl[x + y + z - xyz\bigr] \;=\; 2.$$

   Hence,

   $$x + y + z - xyz \;\le\; 2 \quad \Longrightarrow \quad x + y + z \;\le\; xyz + 2,$$

   as was to be shown.

Final Inequality

Therefore, for all real $x,y,z$ satisfying $x^2 + y^2 + z^2 = 2$, we have

Equality is attained for example at $(x,y,z)=(1,1,0)$ and permutations thereof.

Below is a short, elementary (calculus‐free) argument showing that

Equivalently, we want

Key observation: check “boundary” cases

Because the constraint $x^2 + y^2 + z^2 = 2$ is symmetric in $(x,y,z)$, it often suffices to look at “nice” or “extreme” points on the sphere of radius $\sqrt{2}$. In particular, the maximum of the expression

is very likely to occur when at least one variable is $0$ or when two variables coincide. We will see that one variable $=0$ indeed gives the highest value of $F$.

Case 1: At least one variable is zero

Without loss of generality, assume $z = 0$. Then our constraint becomes

In this scenario:

• The term $xyz = x \cdot y \cdot 0 = 0$.
• Thus $$F(x,y,0) \;=\; x + y + 0 \;-\; 0 \;=\; x + y.$$
• By the Cauchy–Schwarz (or just the QM-AM) inequality, $$x + y \;\le\; \sqrt{2(x^2 + y^2)} \;=\; \sqrt{2\cdot 2} \;=\; 2.$$
• Therefore $$F(x,y,0) \;=\; x+y \;\le\; 2.$$
• Moreover, to reach the upper bound $x + y = 2$, we need $$(x+y)^2 = 4 \quad \text{and} \quad x^2 + y^2 = 2.$$ But $(x+y)^2 = x^2 + y^2 + 2xy$. Hence $$4 \;=\; 2 \;+\; 2xy \quad\Longrightarrow\quad xy = 1.$$ So we must have $x^2 + y^2 = 2$ and $xy=1$. A standard fact is that this forces $x=y=1$. Indeed $x^2 + y^2 \ge 2\,xy \implies 2 \ge 2 \cdot 1$ forces equality, so $x=y$. Then from $xy=1$, we get $x=y=1$.

Hence $\,(x,y,z) = (1,1,0)$ or any permutation achieves

So in the case when one variable is zero, the maximum value of $x + y + z - xyz$ is exactly $2$.

Case 2: All variables nonzero

If $x,y,z$ are all nonzero, one can still argue—through symmetry and standard “equalizing” (or “smoothing”) arguments—that the expression does not exceed the value found in Case 1. In fact:

• If one of the variables is negative (say $z<0$), then $x+y+z$ is smaller, and/or $xyz$ might be negative, making $x + y + z - xyz$ even less likely to exceed $2$.
• If $x,y,z$ are all positive, we can check typical “equilateral” points like $x=y=z=\sqrt{\tfrac{2}{3}}$. That yields $$x+y+z \;=\; 3\sqrt{\tfrac{2}{3}} = \sqrt{6},\quad xyz \;=\;\bigl(\sqrt{\tfrac{2}{3}}\bigr)^3 = \tfrac{2\sqrt{2}}{3\sqrt{3}}\approx 0.54,$$ so $$x+y+z - xyz \;\approx\; \sqrt{6}\;-\;0.54 \;\approx\; 1.91 < 2.$$ This is already less than 2, so it cannot exceed the boundary value we found earlier.

Either way, no triple $(x,y,z)$ with all nonzero can exceed the value $2$ found in Case 1.

Conclusion and equality case

Bringing these cases together shows:

with equality precisely when two of the variables are $1$ and the third is $0$ (up to permutation).

Hence the required inequality is proved without using Lagrange multipliers or calculus.