Okay, so I have this problem where I need to find the largest possible value of |a| + |b| + |c| given that the quadratic function |ax² + bx + c| is bounded by 1 for all x between 0 and 1. Hmm, let me start by understanding the problem better.

First, the function |ax² + bx + c| is always less than or equal to 1 on the interval [0,1]. That means the quadratic doesn't exceed 1 or go below -1 in that interval. The question is asking for the maximum sum of the absolute values of the coefficients a, b, and c. So, I need to maximize |a| + |b| + |c| under the given constraint.

I remember that for polynomials bounded on an interval, there are some results related to Chebyshev polynomials, which are known for minimizing the maximum deviation. But this is about maximizing the sum of the coefficients' absolute values, so maybe a different approach is needed.

Let me think. Since the quadratic is bounded by 1 in absolute value on [0,1], perhaps I can evaluate the quadratic at specific points to get inequalities involving a, b, and c. Then, maybe use those inequalities to find bounds on |a|, |b|, |c| and their sum.

Let me try evaluating the quadratic at x=0, x=1, and also maybe at the vertex of the parabola. Let's see:

At x=0: |c| ≤ 1. So |c| ≤ 1.

At x=1: |a + b + c| ≤ 1. So |a + b + c| ≤ 1.

At the vertex of the parabola, which is at x = -b/(2a). But since x has to be in [0,1], the vertex might be inside or outside the interval. So if the vertex is in [0,1], the value at the vertex is |a*(-b/(2a))² + b*(-b/(2a)) + c| = |(b²)/(4a) - (b²)/(2a) + c| = | - (b²)/(4a) + c |. Hmm, not sure if that's helpful yet.

Alternatively, maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is at least as big as the maximum of |c|, |a + b + c|, and also the maximum at the critical point if it's within [0,1]. So maybe I can set up equations where these maxima are equal to 1, but I need to think about how to maximize |a| + |b| + |c|.

Alternatively, perhaps use Lagrange multipliers to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

But that seems complicated because the constraint is an infinite number of inequalities. Maybe instead, use specific points where the maximum is achieved, and set up equations there.

Wait, if the quadratic is bounded by 1 on [0,1], then the maximum of |ax² + bx + c| occurs at three points: x=0, x=1, and the vertex if it's within [0,1]. So maybe I can set up the equations such that all these points reach the maximum value of 1 or -1.

But since we want to maximize |a| + |b| + |c|, perhaps the optimal case is when the quadratic reaches 1 and -1 at different points in [0,1], which would create the largest possible coefficients.

Alternatively, maybe use the fact that for a quadratic function, the maximum norm on [0,1] can be related to its coefficients. But I'm not sure of the exact relationship.

Wait, maybe use the Markov brothers' inequality or something similar. Markov's inequality relates the maximum of the derivative to the maximum of the function, but not sure if that helps here.

Alternatively, maybe use the fact that for a quadratic polynomial bounded on [0,1], the coefficients can be bounded in terms of the maximum value. But again, not sure.

Alternatively, let's consider that the quadratic function must satisfy |ax² + bx + c| ≤ 1 for all x in [0,1]. Let's write down the inequalities:

For x=0: |c| ≤ 1 ⇒ -1 ≤ c ≤ 1.

For x=1: |a + b + c| ≤ 1 ⇒ -1 ≤ a + b + c ≤ 1.

For x=1/2: |a*(1/4) + b*(1/2) + c| ≤ 1.

But maybe more strategically, let's consider that the maximum of |ax² + bx + c| on [0,1] is equal to the maximum of |c|, |a + b + c|, and the value at the vertex if it's in [0,1]. So perhaps to maximize |a| + |b| + |c|, we need to set up the quadratic such that these maximum values are achieved in a way that the sum |a| + |b| + |c| is maximized.

Alternatively, maybe consider that the maximum of |ax² + bx + c| is equal to the maximum of |c|, |a + b + c|, and | - (b²)/(4a) + c |. So perhaps set these equal to 1 and solve for a, b, c.

But this might get complicated. Let me try to think of an example. Suppose we have a quadratic that reaches 1 at x=0, 1 at x=1, and -1 at x=1/2. Then, we can set up equations:

At x=0: c = 1.

At x=1: a + b + c = 1.

At x=1/2: a*(1/4) + b*(1/2) + c = -1.

So substituting c=1 into the other equations:

a + b + 1 = 1 ⇒ a + b = 0.

a*(1/4) + b*(1/2) + 1 = -1 ⇒ (a/4) + (b/2) = -2.

But since a + b = 0, we can substitute b = -a into the second equation:

(a/4) + (-a/2) = -2 ⇒ (-a/4) = -2 ⇒ a = 8.

Then, b = -8.

So the quadratic is 8x² -8x +1. Let's check the value at x=1/2:

8*(1/4) -8*(1/2) +1 = 2 -4 +1 = -1. Correct.

And the value at x=0 is 1, at x=1 is 8 -8 +1=1. So the quadratic reaches 1 at x=0 and x=1, and -1 at x=1/2. So this quadratic has |a| + |b| + |c| = 8 + 8 +1 =17.

But wait, is this the maximum? Let's check the maximum of |8x² -8x +1| on [0,1]. The maximum is 1, as we saw. So the sum is 17. But the problem says "for all 0 ≤ x ≤1", so the maximum absolute value is 1, and the sum is 17. But is this the maximum possible?

Wait, but maybe we can get a larger sum. Let me try another approach. Suppose that the quadratic touches 1 at x=0 and x=1, and touches -1 at another point in (0,1). Or maybe it's symmetric in some way.

Alternatively, maybe use the fact that the maximum of |ax² + bx + c| is at least |c|, |a + b + c|, and |c - (b²)/(4a)|. So if we can set these equal to 1, then maybe we can solve for a, b, c.

But in the previous example, we had |c| =1, |a + b + c|=1, and |c - (b²)/(4a)|=1. So all three points reach the maximum. So perhaps that configuration gives the maximum sum.

But is there a way to get a higher sum? Let's see.

Suppose that instead of reaching 1 at x=0 and x=1, and -1 at x=1/2, we have the quadratic reaching 1 at x=0, -1 at x=1, and 1 at x=1/2. Let's see what happens.

At x=0: c=1.

At x=1: a + b + c = -1.

At x=1/2: a*(1/4) + b*(1/2) + c =1.

So substituting c=1:

a + b = -2.

(1/4)a + (1/2)b = 0.

Multiply the second equation by 4: a + 2b = 0.

So we have:

a + b = -2.

a + 2b = 0.

Subtract the first equation from the second: b = 2.

Then a = -4.

So the quadratic is -4x² +2x +1. Let's check the maximum on [0,1].

At x=0: 1.

At x=1: -4 +2 +1 = -1.

At x=1/2: -4*(1/4) +2*(1/2) +1 = -1 +1 +1=1. Correct.

So the sum |a| + |b| + |c| is 4 +2 +1=7. Which is less than the previous 17. So the previous case was better.

Alternatively, maybe we can have the quadratic reach 1 at x=0, -1 at x=1/2, and 1 at x=1, but we need to check if that's possible.

But in the previous case, when we set c=1, a + b + c=1, and the vertex at x=1/2 gives -1. That gave a high sum. Maybe that's the maximum.

Alternatively, what if we set the quadratic to reach 1 at x=0, -1 at x=1/4, and 1 at x=1/2. Let's see.

At x=0: c=1.

At x=1/4: a*(1/16) + b*(1/4) +1 = -1 ⇒ (a/16) + (b/4) = -2.

At x=1/2: a*(1/4) + b*(1/2) +1 =1 ⇒ (a/4) + (b/2) =0.

So we have two equations:

1. (a/16) + (b/4) = -2.

2. (a/4) + (b/2) =0.

Multiply the second equation by 4: a + 2b =0 ⇒ a = -2b.

Substitute into the first equation:

(-2b)/16 + (b)/4 = -2 ⇒ (-b/8) + (b/4) = -2 ⇒ (b/8) = -2 ⇒ b= -16.

Then a= -2*(-16)=32.

So the quadratic is 32x² -16x +1.

Check at x=1/4: 32*(1/16) -16*(1/4) +1 =2 -4 +1= -1. Correct.

At x=1/2: 32*(1/4) -16*(1/2) +1 =8 -8 +1=1. Correct.

At x=0: 1. At x=1:32 -16 +1=17. Wait, but the problem states that |ax² +bx +c| ≤1 for all x in [0,1]. But here at x=1, the value is 17, which is way above 1. So that's invalid. So this quadratic is not allowed. So this approach is wrong. So that means, even if we set the quadratic to reach 1 at x=1/4 and x=1/2, but it exceeds 1 at x=1, which violates the condition. So that's not a valid configuration.

Therefore, the previous example where the quadratic reaches 1 at x=0 and x=1, and -1 at x=1/2, gives a quadratic that is within the bounds on [0,1], and the sum |a| + |b| + |c| is 17. But wait, in that case, at x=1, the value is 1, but is there any point where the quadratic exceeds 1 in between?

Wait, let's check the value at x=1/2: -1. Then between x=0 and x=1, the quadratic goes from 1 at x=0 to 1 at x=1, with a dip to -1 in the middle. So the maximum is indeed 1. So that's a valid case. So the sum is 17.

But can we get a higher sum?

Wait, let's try another configuration. Suppose we set the quadratic to reach 1 at x=0, 1 at x=1/3, -1 at x=2/3, and 1 at x=1. Let's see if that's possible.

But this might get too complicated. Alternatively, maybe the maximum sum is 17, as in the first example, and that's the answer. But I need to verify.

Alternatively, let's think of the quadratic as a function on [0,1] with maximum absolute value 1. The maximum of |a| + |b| + |c| would be achieved when the quadratic oscillates between 1 and -1 as much as possible within the interval. The example we had before does that by touching 1 at two points and -1 at the midpoint. So that's a balanced case.

Alternatively, perhaps consider scaling the Chebyshev polynomial. The Chebyshev polynomial of degree 2 is T2(x) = 2x² -1, which oscillates between -1 and 1 on the interval [-1,1]. If we map this to [0,1], we can scale and shift it. Let's see.

Let’s perform a linear transformation to map [-1,1] to [0,1]. Let’s define x = (t +1)/2, where t ∈ [-1,1]. Then substituting into T2(t) = 2t² -1, we get:

T2((2x -1)) = 2*(2x -1)^2 -1 = 2*(4x² -4x +1) -1 =8x² -8x +2 -1=8x² -8x +1.

Wait, that's exactly the quadratic we had earlier: 8x² -8x +1. So this is the Chebyshev polynomial mapped to [0,1]. The maximum of |T2(t)| on [-1,1] is 1, and when mapped to [0,1], the maximum of |8x² -8x +1| is also 1. So this quadratic touches 1 at x=0 and x=1, and -1 at x=1/2, which is exactly the case we considered earlier.

Therefore, this suggests that the maximum sum |a| + |b| + |c| is indeed 17 for this quadratic. Since Chebyshev polynomials are known to minimize the maximum deviation, but in this case, we are maximizing the sum of the coefficients under a boundedness constraint. Since this quadratic achieves the sum of 17 and satisfies the condition, perhaps 17 is the answer.

But wait, let's check another Chebyshev-like polynomial. For example, if we take the polynomial that oscillates three times, but on [0,1], it can only have two extrema. Wait, no. The Chebyshev polynomial of degree n has n+1 extrema in the interval. So on [-1,1], T2(x) has 3 extrema: at -1, between -1 and 1, and at 1. But when mapped to [0,1], it's similar. So the quadratic we have is the Chebyshev polynomial of degree 2 on [0,1], which has maximum at 1 and minimum at -1. Hence, it's the extremal case.

Therefore, the sum of the absolute values of the coefficients is maximized by this polynomial, giving |8| + | -8| + |1| =17.

Hence, the largest possible value is 17.

**Final Answer**
\boxed{17}
Therefore, the largest possible value of $|a| + |b| + |c|$ is $\boxed{17}$.
</think>