If `cos^(-1)(2-x^(2))+sin^(-1)(2-x^(2))+tan^(-1)xge(2pi)/3` have no solution if `xltalpha` or `xgt beta`(for `alphalt beta`) then `{1/("max"(alpha))+("min"(beta))^(2)}=`………

Explanation

To solve the problem step by step, we will analyze the expression given and determine the values of \( \alpha \) and \( \beta \) such that the inequality has no solution for \( x < \alpha \) or \( x > \beta \).

### Step 1: Analyze the given expression
We start with the inequality:
\[ \cos^{-1}(2 - x^2) + \sin^{-1}(2 - x^2) + \tan^{-1}(x) \geq \frac{2\pi}{3} \]

### Step 2: Determine the range of \( 2 - x^2 \)
The functions \( \cos^{-1}(y) \) and \( \sin^{-1}(y) \) are defined for \( y \) in the range \([-1, 1]\). Therefore, we need:
\[ -1 \leq 2 - x^2 \leq 1 \]
This leads to two inequalities:
1. \( 2 - x^2 \geq -1 \) which simplifies to \( x^2 \leq 3 \) or \( |x| \leq \sqrt{3} \).
2. \( 2 - x^2 \leq 1 \) which simplifies to \( x^2 \geq 1 \) or \( |x| \geq 1 \).

Combining these, we find:
\[ 1 \leq |x| \leq \sqrt{3} \]
This means:
\[ x \in [-\sqrt{3}, -1] \cup [1, \sqrt{3}] \]

### Step 3: Simplify the expression using the identity
Using the identity \( \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \), we can rewrite the expression:
\[ \frac{\pi}{2} + \tan^{-1}(x) \geq \frac{2\pi}{3} \]
Subtract \( \frac{\pi}{2} \) from both sides:
\[ \tan^{-1}(x) \geq \frac{2\pi}{3} - \frac{\pi}{2} \]
Calculating the right side:
\[ \frac{2\pi}{3} - \frac{3\pi}{6} = \frac{4\pi - 3\pi}{6} = \frac{\pi}{6} \]
Thus, we have:
\[ \tan^{-1}(x) \geq \frac{\pi}{6} \]

### Step 4: Solve for \( x \)
The inequality \( \tan^{-1}(x) \geq \frac{\pi}{6} \) implies:
\[ x \geq \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

### Step 5: Combine the intervals
From our earlier analysis, we have:
1. \( x \in [-\sqrt{3}, -1] \cup [1, \sqrt{3}] \)
2. \( x \geq \frac{1}{\sqrt{3}} \)

This means the solution set is:
\[ x \in [1, \sqrt{3}] \]
Thus, \( \alpha = \frac{1}{\sqrt{3}} \) and \( \beta = \sqrt{3} \).

### Step 6: Calculate the final expression
We need to find:
\[ \frac{1}{\max(\alpha)} + \frac{1}{\min(\beta)^2} \]
Here, \( \max(\alpha) = \frac{1}{\sqrt{3}} \) and \( \min(\beta) = \sqrt{3} \).

Calculating:
\[ \frac{1}{\frac{1}{\sqrt{3}}} + \frac{1}{(\sqrt{3})^2} = \sqrt{3} + \frac{1}{3} = \sqrt{3} + \frac{1}{3} \]

### Final Answer:
Thus, the final answer is:
\[ \sqrt{3} + \frac{1}{3} \]