Answer :
To find the minimum value of the expression [tex]\( a \sqrt{b^2 + 1} + b \sqrt{c^2 + 1} + c \sqrt{a^2 + 1} \)[/tex] given that [tex]\( a, b, c \in \mathbb{R}^+ \)[/tex] and [tex]\( a b + a c + b c = 1 \)[/tex], follow these steps:
1. Understand the constraint: [tex]\( a b + a c + b c = 1 \)[/tex].
2. Make an initial guess: Because the expressions involve symmetric summations and products, one might guess that the solution could involve [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] being equal. This suggests trying [tex]\(a = b = c\)[/tex].
3. Substitute [tex]\(a = b = c\)[/tex] into the constraint:
[tex]\[ a^2 + a^2 + a^2 = 1 \implies 3a^2 = 1 \implies a^2 = \frac{1}{3} \implies a = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \][/tex]
Thus, we set [tex]\( a = b = c = \frac{1}{\sqrt{3}} \)[/tex].
4. Evaluate the objective function:
[tex]\[ f(a, b, c) = a \sqrt{b^2 + 1} + b \sqrt{c^2 + 1} + c \sqrt{a^2 + 1} \][/tex]
Substituting [tex]\( a = b = c = \frac{1}{\sqrt{3}} \)[/tex]:
[tex]\[ f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} + \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} + \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} \][/tex]
5. Simplify the expressions inside the square roots:
[tex]\[ \left(\frac{1}{\sqrt{3}}\right)^2 + 1 = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
Therefore,
[tex]\[ \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} \][/tex]
6. Evaluate the simplified expression:
[tex]\[ f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 3 \cdot \left(\frac{1}{\sqrt{3}}\right) \cdot \frac{2 \sqrt{3}}{3} \][/tex]
Simplifying further:
[tex]\[ 3 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2 \sqrt{3}}{3} = 3 \cdot \frac{2}{3} = 2 \][/tex]
Thus, the minimum value of the expression [tex]\( a \sqrt{b^2 + 1} + b \sqrt{c^2 + 1} + c \sqrt{a^2 + 1} \)[/tex] given [tex]\( ab + ac + bc = 1 \)[/tex] and [tex]\( a, b, c \in \mathbb{R}^+ \)[/tex] is [tex]\(\boxed{2}\)[/tex].
1. Understand the constraint: [tex]\( a b + a c + b c = 1 \)[/tex].
2. Make an initial guess: Because the expressions involve symmetric summations and products, one might guess that the solution could involve [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] being equal. This suggests trying [tex]\(a = b = c\)[/tex].
3. Substitute [tex]\(a = b = c\)[/tex] into the constraint:
[tex]\[ a^2 + a^2 + a^2 = 1 \implies 3a^2 = 1 \implies a^2 = \frac{1}{3} \implies a = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \][/tex]
Thus, we set [tex]\( a = b = c = \frac{1}{\sqrt{3}} \)[/tex].
4. Evaluate the objective function:
[tex]\[ f(a, b, c) = a \sqrt{b^2 + 1} + b \sqrt{c^2 + 1} + c \sqrt{a^2 + 1} \][/tex]
Substituting [tex]\( a = b = c = \frac{1}{\sqrt{3}} \)[/tex]:
[tex]\[ f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} + \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} + \left(\frac{1}{\sqrt{3}}\right) \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} \][/tex]
5. Simplify the expressions inside the square roots:
[tex]\[ \left(\frac{1}{\sqrt{3}}\right)^2 + 1 = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
Therefore,
[tex]\[ \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} \][/tex]
6. Evaluate the simplified expression:
[tex]\[ f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 3 \cdot \left(\frac{1}{\sqrt{3}}\right) \cdot \frac{2 \sqrt{3}}{3} \][/tex]
Simplifying further:
[tex]\[ 3 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2 \sqrt{3}}{3} = 3 \cdot \frac{2}{3} = 2 \][/tex]
Thus, the minimum value of the expression [tex]\( a \sqrt{b^2 + 1} + b \sqrt{c^2 + 1} + c \sqrt{a^2 + 1} \)[/tex] given [tex]\( ab + ac + bc = 1 \)[/tex] and [tex]\( a, b, c \in \mathbb{R}^+ \)[/tex] is [tex]\(\boxed{2}\)[/tex].
