If `x gt 0,` then the expression `(x ^(100))/( 1 + x + x ^(2) +x ^(3) + ......+ x ^(200))` is always less than or equal to

Answer

To solve the problem, we need to analyze the expression given:

\[ \frac{x^{100}}{1 + x + x^2 + x^3 + \ldots + x^{200}} \]

### Step 1: Simplifying the Denominator
The denominator is a geometric series. The sum of a geometric series can be calculated using the formula:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 1 \), \( r = x \), and the number of terms is \( 201 \) (from \( x^0 \) to \( x^{200} \)).

Thus, the sum of the series is:

\[ 1 + x + x^2 + x^3 + \ldots + x^{200} = \frac{1(1 - x^{201})}{1 - x} = \frac{1 - x^{201}}{1 - x} \]

### Step 2: Substitute the Denominator Back into the Expression
Now, substituting this back into our original expression gives:

\[ \frac{x^{100}}{\frac{1 - x^{201}}{1 - x}} = \frac{x^{100}(1 - x)}{1 - x^{201}} \]

### Step 3: Analyzing the Expression
We want to show that this expression is always less than or equal to some constant. To do this, we will apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step 4: Applying AM-GM Inequality
According to the AM-GM inequality, for any non-negative numbers \( a_1, a_2, \ldots, a_n \):

\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \]

In our case, we can apply AM-GM to the terms in the denominator:

\[ 1, x, x^2, x^3, \ldots, x^{200} \]

The AM of these terms is:

\[ \text{AM} = \frac{1 + x + x^2 + \ldots + x^{200}}{201} \]

The GM of these terms is:

\[ \text{GM} = (1 \cdot x \cdot x^2 \cdots x^{200})^{\frac{1}{201}} = x^{\frac{0 + 1 + 2 + \ldots + 200}{201}} = x^{\frac{200 \cdot 201 / 2}{201}} = x^{100} \]

By AM-GM, we have:

\[ \frac{1 + x + x^2 + \ldots + x^{200}}{201} \geq x^{100} \]

### Step 5: Rearranging the Inequality
From the AM-GM inequality, we can rearrange it to find:

\[ 1 + x + x^2 + \ldots + x^{200} \geq 201 x^{100} \]

### Step 6: Final Expression
Now substituting this back into our expression, we get:

\[ \frac{x^{100}}{1 + x + x^2 + \ldots + x^{200}} \leq \frac{x^{100}}{201 x^{100}} = \frac{1}{201} \]

Thus, we conclude that:

\[ \frac{x^{100}}{1 + x + x^2 + \ldots + x^{200}} \leq \frac{1}{201} \]

### Final Answer
The expression is always less than or equal to \( \frac{1}{201} \).

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