Question

Assume that \(x+y+z=1\) with \(x \geq 0\), \(y \geq 0,\) and \(z \geq 0\). a. Find the maximum and minimum values of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)\) b. Find the maximum and minimum values of \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z})\)

Short answer

The minimum and maximum values for the given expressions are: a) For \((1+x^2)(1+y^2)(1+z^2)\), the minimum value is \(\frac{64}{27}\) and there is no maximum value. b) For \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z})\), the minimum value is \(8\) and the maximum value is \(16\).

Step-by-step solution

Exploit symmetry

By symmetry, we can assume that \(x \leq y \leq z\) as it will not affect the final answer. So, we can focus on the range \(0 \leq x \leq y \leq z \leq 1\).

Apply AM-GM inequality on each factor

Applying AM-GM inequality on \(1+x^2\), we get: $$ \frac{1+x^2}{2} \geq \sqrt{(1)(x^2)} \Rightarrow (1+x^2) \geq 2x $$ Similarly, we have: $$ (1+y^2) \geq 2y \\ (1+z^2) \geq 2z $$

Combine the inequalities

Multiply the inequalities obtained in Step 2 to get: $$ (1+x^2)(1+y^2)(1+z^2) \geq 8xyz $$

Apply AM-GM inequality on \(x+y+z\)

Recall that \(x+y+z=1\). Applying AM-GM inequality on \(x,y,z\): $$ \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \Rightarrow 1 \geq 3\sqrt[3]{xyz} \Rightarrow \left(\frac{1}{3}\right)^3 \geq xyz $$

Find the minimum value

Combining the results from Steps 3 and 4, we get \((1+x^2)(1+y^2)(1+z^2) \geq 64\left(\frac{1}{3}\right)^3\). So the minimum value is \(\boxed{\frac{64}{27}}\), which occurs when \(x=y=z=\frac{1}{3}\).

There is no maximum value

Note that \((1+x^2)(1+y^2)(1+z^2)\) can be arbitrarily large as the terms within the brackets can grow indefinitely. Therefore, there is no maximum value for the given expression. #b. Find the maximum and minimum values of \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z})\)

Apply AM-GM inequality on each factor

Applying AM-GM inequality on \(1+\sqrt{x}\), we get: $$ \frac{1+\sqrt{x}}{2} \geq \sqrt{(1)(\sqrt{x})} \Rightarrow (1+\sqrt{x}) \geq 2\sqrt{x} $$ Similarly, we have: $$ (1+\sqrt{y}) \geq 2\sqrt{y} \\ (1+\sqrt{z}) \geq 2\sqrt{z} $$

Combine the inequalities

Multiply the inequalities obtained in Step 1 to get: $$ (1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z}) \geq 8\sqrt[3]{xyz} $$

Apply AM-GM inequality on \(x+y+z\)

As we did in part a, apply the AM-GM inequality on \(x,y,z\): $$ \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \Rightarrow 1 \geq 3\sqrt[3]{xyz} \Rightarrow \left(\frac{1}{3}\right)^3 \geq xyz $$

Find the minimum value

Combining the results from Steps 2 and 3, we get \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z}) \geq 64\left(\frac{1}{3}\right)^3\). So the minimum value is \(\boxed{8}\), which occurs when \(x=y=z=\frac{1}{3}\).

Express the objective function in terms of \(y\) and \(z\)

Since \(x+y+z=1\), we have \(x=1-y-z\). The objective function becomes: \((1+\sqrt{1-y-z})(1+\sqrt{y})(1+\sqrt{z})\). Let's rewrite the expression in terms of \(y\) and \(z\), \(f(y,z)=(2-\sqrt{y+z})(2+\sqrt{y})(2+\sqrt{z})\). We want to find the maximum value of this function.

Notice the bound of f(y,z)

Notice that, by AM-GM inequality, \(\sqrt{y+z} \leq \frac{y+z}{2}\leq\frac{1}{2}\). Therefore, \(\ -\frac{1}{2} \leq \sqrt{y+z} - 1 \leq 0 \), and we can rewrite f(y,z) as \((1 - (\sqrt{y+z} - 1))(1+\sqrt{y})(1+\sqrt{z})\). Since \(\ -\frac{1}{2} \leq \sqrt{y+z} - 1 \leq 0 \), we know \(1 - (\sqrt{y+z} - 1) \leq 2\). Hence, the bound of f(y,z) would be \(2(1+\sqrt{y})(1+\sqrt{z})\).

Find the maximum value

We know the bound of \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z})\) is \(2(1+\sqrt{y})(1+\sqrt{z})\). However, \((1+\sqrt{x})(1+\sqrt{y})(1+\sqrt{z}) \leq 2(1+\sqrt{y})(1+\sqrt{z})\) is not necessarily true. To find the exact maximum value, we must show when equality holds. Notice that the equality holds in Step 1 only if \(1=\sqrt{x}\) or \(x=0\). Since x+y+z=1, we must have \(1=\sqrt{y}\) and \(1=\sqrt{z}\) or \(y=z=1\). Therefore, the maximum value is \(2(1+1)(1+1)=\boxed{16}\), when \(x=0\), \(y=z=1\).

Key concepts

Symmetry in Equations

Symmetry in equations is a powerful mathematical tool used to simplify problems. When dealing with variables in an equation, assuming a certain order of the variables can help us analyze the situation more easily. Here, in the exercise, because the problem is symmetric in terms of the variables \(x, y, \) and \(z\), we assume \( x \leq y \leq z \). This assumption doesn't change the overall outcome but allows us to focus on the range \(0 \leq x \leq y \leq z \leq 1\).

By exploiting this symmetry, we're able to reduce complexity, as we don't have to consider every possible arrangement of \(x, y, \) and \(z\). For similar problems, always look for symmetry in the equation or conditions. This property can save time and simplify calculations. Not only does it help with algebraic manipulations, but it also provides insights into the problem's structure and potential solutions.

• Symmetry reduces the number of cases to consider.
• Assumptions like \(x \leq y \leq z\) don't affect the final results.
• Look for symmetry in expressions and boundaries.

Identifying symmetry can often reveal underlying patterns and connections in the problem that are not immediately obvious from the original formulation.

Extreme Value Theorem

The Extreme Value Theorem is a fundamental principle that states a continuous function over a closed interval must have both a maximum and minimum value. However, in some cases, like in our exercise, the function can grow indefinitely.

For this exercise, the expression \((1+x^2)(1+y^2)(1+z^2)\) does not have a definite maximum value within the given constraint \(x+y+z=1\). As any of the variables approach zero, another can reach closer to one, leading to very large values of the terms \(1+y^2\) or \(1+z^2\). Hence, the value of\((1+x^2)(1+y^2)(1+z^2)\) can become arbitrarily large, leaving it technically without a maximum value.

• The function must be continuous over a closed interval to apply the theorem directly.
• Extreme values might not exist if the function grows indefinitely.

In scenarios involving mathematical optimization or understanding function behavior, the Extreme Value Theorem highlights the necessity of defining boundaries or constraints properly. Without these, the function's behavior at the extreme ends might lead to undefined or infinite outcomes.

Optimization in Calculus

Optimization in calculus involves finding the maximum or minimum values of a function, often subject to constraints. In our problem, leveraging the AM-GM inequality allows us to estimate the minimum value of the given expressions.

Applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality helps to identify critical points and boundary values which can give potential minima or maxima under the constraint \(x+y+z=1\). By breaking down complex expressions into simpler components, like \(1+x^2\), \(1+y^2\), and \(1+z^2\), we can establish lower bounds based on their arithmetic and geometric means.

• The AM-GM inequality is useful for deriving potential extremal values in optimization.
• Identifying critical points is essential when constraints are involved.
• AM-GM transforms complex expressions into manageable algebraic inequalities.

In calculus and real-world problems, applying these methods reveals critical insights, guiding us to solutions that satisfy given conditions while optimizing the desired outcome, such as minimizing cost or maximizing efficiency.