Write SQL query to solve given problem: Among the establishments that failed the inspection in February 2010, list the names of the employees with a salary greater than 70% of the average salary of all employees.. Keep the solution inside sql tag ```sql [SQL-Query] ```
SELECT DISTINCT T1.employee_id FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T2.results = 'Fail' AND strftime('%Y-%m', T2.inspection_date) = '2010-02' AND T1.salary > 0.7 * ( SELECT AVG(salary) FROM employee )