Algebraic Interval Determination

This problem asks us to find the range or interval in which the expression \({\rm{ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\) lies, given the condition that \({\rm{a}}^2{\rm{\;}} + {\rm{\;}}{{\rm{b}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{c}}^2}{\rm{\;}} = {\rm{\;}}1\). To solve this, we will use fundamental algebraic identities involving squares of real numbers.

Understanding the Key Algebraic Identity

A crucial algebraic identity relates the sum of squares and the sum of products:

For any real numbers \(a\), \(b\), and \(c\):

\((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac)\)

We are given \(a^2 + b^2 + c^2 = 1\). Let's substitute this into the identity:

\((a+b+c)^2 = 1 + 2(ab+bc+ac)\)

Finding the Lower Bound for \({\rm{ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\)

Since \(a\), \(b\), and \(c\) are real numbers, the square of their sum, \((a+b+c)^2\), must always be greater than or equal to zero.

So, we have:

\((a+b+c)^2 \ge 0\)

Substituting the expression we derived:

\(1 + 2(ab+bc+ac) \ge 0\)

Now, let's solve for \(ab+bc+ac\):

\(2(ab+bc+ac) \ge -1\)

\(ab+bc+ac \ge -\frac{1}{2}\)

To confirm this lower bound is achievable, consider the values \(a = \frac{1}{\sqrt{2}}\), \(b = -\frac{1}{\sqrt{2}}\), and \(c = 0\).

• Checking the condition: \(a^2 + b^2 + c^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + 0^2 = \frac{1}{2} + \frac{1}{2} + 0 = 1\). The condition is satisfied.
• Calculating \(ab+bc+ac\): \(ab+bc+ac = \left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{1}{\sqrt{2}}\right)(0) + (0)\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2} + 0 + 0 = -\frac{1}{2}\).

Since we found values for \(a, b, c\) that satisfy the condition and result in \(ab+bc+ac = -\frac{1}{2}\), this lower bound is indeed attainable.

Finding the Upper Bound for \({\rm{ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\)

Another important algebraic property for real numbers is that the sum of squares of differences is always non-negative:

\((a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0\)

Expanding this expression:

\((a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) \ge 0\)

\(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac \ge 0\)

Divide by 2:

\(a^2 + b^2 + c^2 - (ab+bc+ac) \ge 0\)

Rearrange the inequality:

\(a^2 + b^2 + c^2 \ge ab+bc+ac\)

We are given \(a^2 + b^2 + c^2 = 1\). Substituting this value:

\(1 \ge ab+bc+ac\)

So, \(ab+bc+ac \le 1\).

To confirm this upper bound is achievable, consider the values \(a = b = c = \frac{1}{\sqrt{3}}\).

• Checking the condition: \(a^2 + b^2 + c^2 = \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\). The condition is satisfied.
• Calculating \(ab+bc+ac\): \(ab+bc+ac = \left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right) + \left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right) + \left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\).

Since we found values for \(a, b, c\) that satisfy the condition and result in \(ab+bc+ac = 1\), this upper bound is indeed attainable.

Conclusion on the Interval

Combining both the lower bound and the upper bound, we find that the expression \(ab+bc+ac\) must lie within the interval:

\(-\frac{1}{2} \le ab+bc+ac \le 1\)

In interval notation, this is written as \(\left[ { - \frac{1}{2},{\rm{\;}}1} \right]\).

Let's summarize the derivation:

Therefore, the interval for \({\rm{ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\) is \(\left[ { - \frac{1}{2},{\rm{\;}}1} \right]\).