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           ☆1 inequalities

[ 1/1Inequality]
    Assignment:Find the solution set of inequality 0.5 <= (3a^2+2a+sqrt(9a^4+6a^3+7a^2+4a+1)+1)/(3a^2) <= 2 .
    Question type: Inequality
    Solution:
    The inequality can be reduced to 2 inequalities：
        0.5 <= ( 3 * a ^ 2 + 2 * a + sqrt ( 9 * a ^ 4 + 6 * a ^ 3 + 7 * a ^ 2 + 4 * a + 1 ) + 1 ) / ( 3 * a ^ 2 )         (1)
         ( 3 * a ^ 2 + 2 * a + sqrt ( 9 * a ^ 4 + 6 * a ^ 3 + 7 * a ^ 2 + 4 * a + 1 ) + 1 ) / ( 3 * a ^ 2 ) <= 2         (2)
        From the definition field of √
         9 * x ^ 4 + 6 * x ^ 3 + 7 * x ^ 2 + 4 * x + 1 ≥ 0        (3 )
        From the definition field of divisor
         3 * x ^ 2 ≠ 0        (4 )

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    From inequality(1)：
         a ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!
    From inequality(2)：
         a ≤ -1/2
    From inequality(3)：
         a ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!
    From inequality(4)：
         a < 0 或  a ＞ 0

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    From inequalities (1) and (2)
         a ≤ -1/2    (5)
    From inequalities (3) and (5)
         a ≤ -1/2    (6)
    From inequalities (4) and (6)
         a ≤ -1/2    (7)

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    The final solution set is :
         a ≤ -1/2

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