Okay, let's try to figure out the smallest possible value of \( a \) such that all roots of the cubic equation \( x^3 + a x^2 + a x + 1 = 0 \) are real. Hmm, I remember that for polynomials, having all real roots can sometimes be related to the discriminant, but I'm not sure if that's the easiest way here. Maybe there's another approach.

First, let me write down the equation again to make sure I have it right:

\( x^3 + a x^2 + a x + 1 = 0 \).

I notice that the coefficients are symmetric in a way. The coefficients of \( x^3 \) and the constant term are both 1, and the coefficients of \( x^2 \) and \( x \) are both \( a \). Maybe this symmetry can be exploited somehow. Perhaps there's a substitution or a factorization that can simplify the problem.

Wait, if I let \( y = x + \frac{1}{x} \), maybe? Let me check. If I divide both sides by \( x \) (assuming \( x \neq 0 \)), we get:

\( x^2 + a x + a + \frac{1}{x} = 0 \).

But that doesn't seem to help much. Alternatively, maybe factoring by grouping. Let's see:

\( x^3 + a x^2 + a x + 1 = (x^3 + 1) + a x (x + 1) \).

Hmm, \( x^3 + 1 \) can be factored as \( (x + 1)(x^2 - x + 1) \), so:

\( (x + 1)(x^2 - x + 1) + a x (x + 1) = (x + 1)(x^2 - x + 1 + a x) \).

So, the equation factors as \( (x + 1)(x^2 + (a - 1)x + 1) = 0 \). Therefore, one root is \( x = -1 \), and the other two roots are solutions to \( x^2 + (a - 1)x + 1 = 0 \).

Since we want all roots to be real, the quadratic \( x^2 + (a - 1)x + 1 = 0 \) must have real roots. The discriminant of this quadratic must be non-negative. The discriminant \( D \) is:

\( D = (a - 1)^2 - 4 \times 1 \times 1 = (a - 1)^2 - 4 \).

For the quadratic to have real roots, we need \( D \geq 0 \):

\( (a - 1)^2 - 4 \geq 0 \)

\( (a - 1)^2 \geq 4 \)

Taking square roots:

\( |a - 1| \geq 2 \)

Which gives two cases:

1. \( a - 1 \geq 2 \) ⇒ \( a \geq 3 \)
2. \( a - 1 \leq -2 \) ⇒ \( a \leq -1 \)

But the problem states that \( a \) is a positive real number. So, the second case \( a \leq -1 \) is invalid. Therefore, the smallest possible value of \( a \) is 3.

Wait, but let me check if this makes sense. If \( a = 3 \), then the quadratic becomes \( x^2 + 2x + 1 = 0 \), which factors as \( (x + 1)^2 = 0 \), so the roots are \( x = -1 \) (double root) and \( x = -1 \). Therefore, all roots are real. But if \( a \) is slightly less than 3, say \( a = 2.9 \), then the discriminant would be \( (2.9 - 1)^2 - 4 = (1.9)^2 - 4 = 3.61 - 4 = -0.39 \), which is negative, so the quadratic would have complex roots. Hence, the cubic equation would have one real root and two complex conjugate roots. Therefore, \( a = 3 \) is indeed the minimal value where all roots are real.

But wait, another thought: maybe there's a different substitution or factorization that allows for a different minimum. Let me think again. Since we factored the cubic into \( (x + 1)(x^2 + (a - 1)x + 1) \), the roots are \( x = -1 \) and the roots of the quadratic. So, as long as the quadratic has real roots, all roots of the cubic are real. Therefore, the discriminant condition is necessary and sufficient.

Hence, the minimal \( a \) is 3. Wait, but hold on. Let me verify this by plugging \( a = 3 \) back into the original equation:

\( x^3 + 3x^2 + 3x + 1 = 0 \).

Factorizing:

\( (x + 1)(x^2 + 2x + 1) = 0 \)

Which is \( (x + 1)^3 = 0 \), so the triple root at \( x = -1 \). So all roots are real. If \( a \) were less than 3, say \( a = 2 \), then the quadratic would be \( x^2 + x + 1 \), discriminant \( 1 - 4 = -3 \), so complex roots. Therefore, the cubic would have one real root and two complex roots. Therefore, 3 is indeed the minimal value.

But wait, just to make sure there's no other way to have all roots real with a smaller \( a \). Suppose we consider the cubic equation \( x^3 + a x^2 + a x + 1 = 0 \). Let's analyze the behavior of the function \( f(x) = x^3 + a x^2 + a x + 1 \).

For the cubic to have all real roots, its graph must cross the x-axis three times. The function tends to \( +\infty \) as \( x \to \infty \) and \( -\infty \) as \( x \to -\infty \). The critical points are where \( f'(x) = 0 \). Let's compute the derivative:

\( f'(x) = 3x^2 + 2a x + a \).

Set this equal to zero:

\( 3x^2 + 2a x + a = 0 \).

The discriminant of this quadratic is \( (2a)^2 - 4 \times 3 \times a = 4a^2 - 12a = 4a(a - 3) \).

For real critical points, the discriminant must be non-negative:

\( 4a(a - 3) \geq 0 \).

Since \( a \) is positive, this inequality holds when \( a \geq 3 \). Therefore, if \( a < 3 \), the derivative has no real roots, meaning the function \( f(x) \) is strictly increasing (since the leading coefficient is positive). Therefore, \( f(x) \) would have only one real root, and two complex roots. But when \( a = 3 \), the derivative has a double root at \( x = -1 \), which is the point where the function has a horizontal tangent. So, the function touches the x-axis there, leading to a multiple root.

Therefore, this confirms that \( a = 3 \) is indeed the minimal value where the cubic has all real roots (with a triple root at \( x = -1 \)).

Therefore, the smallest possible value of \( a \) is 3.

**Final Answer**
The smallest possible value of \( a \) is \boxed{3}.
Thus, the smallest possible value of \( a \) is $\boxed{3}$.
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