求证：恒正二次多项式满足$f(x)+f'(x)+f''(x)＞0$（需严谨证明）

Hey there! I see you've already tested this with numerical examples, so let's lock in a rigorous, step-by-step proof. We'll cover two solid approaches to make sure this makes sense from multiple angles.

Approach 1: Direct Expansion and Discriminant Analysis #

Since ( f(x) ) is a quadratic polynomial that's always positive for all real ( x ), we can write it in standard form:
[ f(x) = ax^2 + bx + c ]
where:

• ( a > 0 ) (the parabola opens upwards)
• The discriminant ( \Delta = b^2 - 4ac < 0 ) (no real roots, so the parabola never touches or crosses the x-axis)

First, calculate the first and second derivatives:

• ( f'(x) = 2ax + b )
• ( f''(x) = 2a )

Add them all together to get the expression we need to analyze:
[
\begin{align*}
f(x) + f'(x) + f''(x) &= ax^2 + bx + c + 2ax + b + 2a \
&= ax^2 + (b + 2a)x + (c + b + 2a)
\end{align*}
]

This is another quadratic polynomial. To prove it's always positive, we just need two things:

1. The leading coefficient ( a > 0 ) (which we already know is true)
2. Its discriminant is negative (so it has no real roots and stays above the x-axis)

Compute the discriminant ( \Delta' ) of this new quadratic:
[
\begin{align*}
\Delta' &= (b + 2a)^2 - 4a(c + b + 2a) \
&= b^2 + 4ab + 4a^2 - 4ac - 4ab - 8a^2 \
&= b^2 - 4ac - 4a^2
\end{align*}
]

We know ( b^2 - 4ac = \Delta < 0 ) (from the original polynomial's positivity), and ( -4a^2 < 0 ) (since ( a > 0 )). Adding two negative numbers gives ( \Delta' < 0 ).

Since the new quadratic has a positive leading coefficient and no real roots, it's always positive for all real ( x ).

Approach 2: Using an Auxiliary Function #

Let's construct a helper function to simplify the analysis:
[ g(x) = e^x f(x) ]

Use the product rule to find its first and second derivatives:

• ( g'(x) = e^x f(x) + e^x f'(x) = e^x\left(f(x) + f'(x)\right) )
• ( g''(x) = e^x\left(f(x) + f'(x)\right) + e^x\left(f'(x) + f''(x)\right) = e^x\left(f(x) + f'(x) + f''(x)\right) )

Since ( e^x > 0 ) for all real ( x ), the sign of ( g''(x) ) is exactly the same as the sign of ( f(x) + f'(x) + f''(x) ).

Substitute ( f(x) = ax^2 + bx + c ) into ( g''(x) ):
[ g''(x) = e^x\left(ax2 + (b + 2a)x + (c + b + 2a)\right) ]

From Approach 1, we already know the quadratic inside the parentheses is always positive, and ( e^x ) is always positive. Multiplying two positive values gives ( g''(x) > 0 ) for all ( x ), which means ( f(x) + f'(x) + f''(x) > 0 ) for all real ( x ).

Alternatively, we could look at the minimum value of the quadratic ( f(x) + f'(x) + f''(x) ). Since it opens upwards, its minimum occurs at ( x = -\frac{b + 2a}{2a} = -\frac{b}{2a} - 1 ). Substituting this into the expression simplifies to ( \left(c - \frac{b^2}{4a}\right) + a ):

• ( c - \frac{b^2}{4a} ) is the minimum value of ( f(x) ), which is positive (since ( f(x) > 0 ) everywhere)
• ( a > 0 )

Adding two positive numbers gives a positive result, so the minimum value is positive—meaning the entire expression is always positive.

内容的提问来源于stack exchange，提问作者Tilak Madichetti