Question 1018968
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Solve the polynomial inequality. Express the solution set in interval notation.

4x^2 + 1 is greater than or equal to 4x
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{{{4x^2 + 1}}} >= {{{4x}}}.    (1)

This inequality is equivalent to

{{{4x^2 - 4x + 1}}} >= {{{0}}}.

Left side is {{{(2x-1)^2}}}. So the inequality takes the form

{{{(2x-1)^2}}} >= {{{0}}}.

The last inequality is true for any x, because its left side is always non-negative as the square of real number (2x-1).

Therefore, the original equation (1) is valid for all real numbers x.

Thus the set of solutions to inequality (1) is all real numbers, the entire number line ({{{-infinity}}}, {{{infinity}}}).
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