Equation `x^4+ax^3+bx^2+cx+1=0` has real roots (a,b,c are non-negative).
Minimum non-negative real value of b is (a) 12 (b) 15 (c) 6 (d) 10

Options

A. 12
B. 15
C. 6
D. 10

Explanation

To solve the equation \( x^4 + ax^3 + bx^2 + cx + 1 = 0 \) with real roots, where \( a, b, c \) are non-negative, we need to determine the minimum non-negative real value of \( b \).

### Step-by-step Solution:

1. **Understanding the Roots**:
Since the polynomial has degree 4, it has 4 roots. Let's denote the roots as \( x_1, x_2, x_3, x_4 \). Given that \( a, b, c \) are non-negative, we conclude that the roots must be negative. This is because if the roots were positive, the polynomial would not be able to equal zero (as all terms would be positive).

2. **Using Vieta's Formulas**:
According to Vieta's formulas for a polynomial \( x^4 + ax^3 + bx^2 + cx + 1 = 0 \):
- The sum of the roots \( x_1 + x_2 + x_3 + x_4 = -a \) (which is positive since \( a \geq 0 \)).
- The sum of the products of the roots taken two at a time \( x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = b \).
- The product of the roots \( x_1 x_2 x_3 x_4 = 1 \).

3. **Applying AM-GM Inequality**:
We apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the sums of the products of the roots:
\[ \frac{x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4}{6} \geq \sqrt[6]{(x_1x_2x_3x_4)^3} \]
Here, the left-hand side is equal to \( \frac{b}{6} \) and the right-hand side simplifies to:
\[ \sqrt[6]{(1)^3} = 1 \]
Therefore, we have:
\[ \frac{b}{6} \geq 1 \]

4. **Solving for \( b \)**:
From the inequality \( \frac{b}{6} \geq 1 \), we can multiply both sides by 6 to find:
\[ b \geq 6 \]

5. **Finding the Minimum Value**:
The minimum non-negative real value of \( b \) that satisfies this inequality is \( b = 6 \).

6. **Conclusion**:
We check the options provided:
- (a) 12
- (b) 15
- (c) 6
- (d) 10

The correct answer is \( \boxed{6} \).