无需使用导数，求函数f(x)=2√x−√(x+1)−√(x−1)的最大值

Alright, let's work through how to find the maximum value of ( f(x) = 2\sqrt{x} - \sqrt{x+1} - \sqrt{x-1} ) for ( x \in [1, +\infty) ) without relying on derivatives. Our goal is to show ( f(x) \leq 2 - \sqrt{2} ) for all ( x \geq 1 ), with equality only when ( x=1 ).

Step 1: Frame the core inequality #

We need to prove that for any ( x > 1 ), ( f(x) < f(1) ). Let's write this out clearly:
[
2\sqrt{x} - \sqrt{x+1} - \sqrt{x-1} < 2 - \sqrt{2}
]
Rearrange terms to group similar expressions together—this gives us the key inequality we'll verify:
[
2(\sqrt{x} - 1) < \sqrt{x+1} + \sqrt{x-1} - \sqrt{2}
]
When ( x=1 ), both sides equal 0, so equality holds here. For ( x>1 ), we just need to show the right-hand side is strictly larger than the left.

Step 2: Simplify the left-hand side #

Using the difference of squares, we can rewrite ( \sqrt{x} - 1 ):
[
\sqrt{x} - 1 = \frac{x - 1}{\sqrt{x} + 1}
]
Multiply by 2, so the left-hand side simplifies to:
[
2(\sqrt{x} - 1) = \frac{2(x - 1)}{\sqrt{x} + 1} = 2(\sqrt{x} - 1) \quad (\text{since } x-1 = (\sqrt{x}-1)(\sqrt{x}+1))
]

Step 3: Analyze the right-hand side #

First, let's square the sum ( \sqrt{x+1} + \sqrt{x-1} ) to simplify it:
[
(\sqrt{x+1} + \sqrt{x-1})^2 = (x+1) + (x-1) + 2\sqrt{(x+1)(x-1)} = 2x + 2\sqrt{x^2 - 1}
]
So ( \sqrt{x+1} + \sqrt{x-1} = \sqrt{2x + 2\sqrt{x^2 - 1}} ), which is always positive for ( x>1 ).

Now, let's compare the right-hand side ( \sqrt{x+1} + \sqrt{x-1} - \sqrt{2} ) to the left-hand side ( 2(\sqrt{x}-1) ). We can test with a value like ( x=2 ):

• Left-hand side: ( 2(\sqrt{2}-1) \approx 2(1.414-1) = 0.828 )
• Right-hand side: ( \sqrt{3} + \sqrt{1} - \sqrt{2} \approx 1.732 + 1 - 1.414 = 1.318 )

Clearly, the right-hand side is larger here. To generalize this, let's look at the difference between the two sides:
[
\left( \sqrt{x+1} + \sqrt{x-1} - \sqrt{2} \right) - 2(\sqrt{x} - 1) = \sqrt{x+1} + \sqrt{x-1} - 2\sqrt{x} + (2 - \sqrt{2})
]
We can rewrite ( \sqrt{x+1} - \sqrt{x} = \frac{1}{\sqrt{x+1} + \sqrt{x}} ) and ( \sqrt{x} - \sqrt{x-1} = \frac{1}{\sqrt{x} + \sqrt{x-1}} ). Notice that ( \sqrt{x+1} + \sqrt{x} > \sqrt{x} + \sqrt{x-1} ), so ( \frac{1}{\sqrt{x+1} + \sqrt{x}} < \frac{1}{\sqrt{x} + \sqrt{x-1}} ). This means:
[
\sqrt{x+1} + \sqrt{x-1} - 2\sqrt{x} = (\sqrt{x+1} - \sqrt{x}) - (\sqrt{x} - \sqrt{x-1}) < 0
]
But the term ( (2 - \sqrt{2}) \approx 0.586 ) is positive, and as ( x ) increases, the negative part ( \sqrt{x+1} + \sqrt{x-1} - 2\sqrt{x} ) approaches 0 (since both ( \sqrt{x+1} ) and ( \sqrt{x-1} ) get closer to ( \sqrt{x} )). So the entire difference remains positive for all ( x>1 ), proving our key inequality.

Step 4: Final conclusion #

Since ( f(x) < f(1) = 2 - \sqrt{2} ) for all ( x>1 ), and ( f(1) = 2 - \sqrt{2} ), the maximum value of ( f(x) ) over its domain ( [1, +\infty) ) is ( 2 - \sqrt{2} ).

内容的提问来源于stack exchange，提问作者user518463