Solve 1/2x-[(x+1/2)^{2}-(x-1/2)^{2}]+1/2x^{2}geq[(frac{x+1}{2})^2+(frac{x-1}{2})^2]^2-frac{1}{4}x^4

$$\frac { 1 } { 2 } x - [ ( \frac { x + 1 } { 2 } ) ^ { 2 } - ( \frac { x - 1 } { 2 } ) ^ { 2 } ] + \frac { 1 } { 2 } x ^ { 2 } \geq [ ( \frac { x + 1 } { 2 } ) ^ { 2 } + ( \frac { x - 1 } { 2 } ) ^ { 2 } ] ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$$

Solve for x

$$x\leq -\frac{1}{2}$$

Short Solution Steps

To raise \frac{x+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1}{2}x-\left(\frac{\left(x+1\right)^{2}}{2^{2}}-\left(\frac{x-1}{2}\right)^{2}\right)+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

To raise \frac{x-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1}{2}x-\left(\frac{\left(x+1\right)^{2}}{2^{2}}-\frac{\left(x-1\right)^{2}}{2^{2}}\right)+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\frac{1}{2}x-\left(\frac{\left(x+1\right)^{2}}{2^{2}}-\frac{x^{2}-2x+1}{2^{2}}\right)+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{2}x-\left(\frac{\left(x+1\right)^{2}}{2^{2}}-\frac{x^{2}-2x+1}{4}\right)+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{1}{2}x-\left(\frac{\left(x+1\right)^{2}}{4}-\frac{x^{2}-2x+1}{4}\right)+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Since \frac{\left(x+1\right)^{2}}{4} and \frac{x^{2}-2x+1}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x-\frac{\left(x+1\right)^{2}-\left(x^{2}-2x+1\right)}{4}+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Do the multiplications in \left(x+1\right)^{2}-\left(x^{2}-2x+1\right).

$$\frac{1}{2}x-\frac{x^{2}+2x+1-x^{2}+2x-1}{4}+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Combine like terms in x^{2}+2x+1-x^{2}+2x-1.

$$\frac{1}{2}x-\frac{4x}{4}+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Cancel out 4 and 4.

$$\frac{1}{2}x-x+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

Combine \frac{1}{2}x and -x to get -\frac{1}{2}x.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\left(\frac{x+1}{2}\right)^{2}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

To raise \frac{x+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{\left(x+1\right)^{2}}{2^{2}}+\left(\frac{x-1}{2}\right)^{2}\right)^{2}-\frac{1}{4}x^{4}$$

To raise \frac{x-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{\left(x+1\right)^{2}}{2^{2}}+\frac{\left(x-1\right)^{2}}{2^{2}}\right)^{2}-\frac{1}{4}x^{4}$$

Since \frac{\left(x+1\right)^{2}}{2^{2}} and \frac{\left(x-1\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{\left(x+1\right)^{2}+\left(x-1\right)^{2}}{2^{2}}\right)^{2}-\frac{1}{4}x^{4}$$

Do the multiplications in \left(x+1\right)^{2}+\left(x-1\right)^{2}.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{x^{2}+2x+1+x^{2}-2x+1}{2^{2}}\right)^{2}-\frac{1}{4}x^{4}$$

Combine like terms in x^{2}+2x+1+x^{2}-2x+1.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{2x^{2}+2}{2^{2}}\right)^{2}-\frac{1}{4}x^{4}$$

Factor the expressions that are not already factored in \frac{2x^{2}+2}{2^{2}}.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{2\left(x^{2}+1\right)}{2^{2}}\right)^{2}-\frac{1}{4}x^{4}$$

Cancel out 2 in both numerator and denominator.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \left(\frac{x^{2}+1}{2}\right)^{2}-\frac{1}{4}x^{4}$$

To raise \frac{x^{2}+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{\left(x^{2}+1\right)^{2}}{2^{2}}-\frac{1}{4}x^{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+1\right)^{2}.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{\left(x^{2}\right)^{2}+2x^{2}+1}{2^{2}}-\frac{1}{4}x^{4}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{x^{4}+2x^{2}+1}{2^{2}}-\frac{1}{4}x^{4}$$

Calculate 2 to the power of 2 and get 4.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{x^{4}+2x^{2}+1}{4}-\frac{1}{4}x^{4}$$

Divide each term of x^{4}+2x^{2}+1 by 4 to get \frac{1}{4}x^{4}+\frac{1}{2}x^{2}+\frac{1}{4}.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{1}{4}x^{4}+\frac{1}{2}x^{2}+\frac{1}{4}-\frac{1}{4}x^{4}$$

Combine \frac{1}{4}x^{4} and -\frac{1}{4}x^{4} to get 0.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}\geq \frac{1}{2}x^{2}+\frac{1}{4}$$

Subtract \frac{1}{2}x^{2} from both sides.

$$-\frac{1}{2}x+\frac{1}{2}x^{2}-\frac{1}{2}x^{2}\geq \frac{1}{4}$$

Combine \frac{1}{2}x^{2} and -\frac{1}{2}x^{2} to get 0.

$$-\frac{1}{2}x\geq \frac{1}{4}$$

Multiply both sides by -2, the reciprocal of -\frac{1}{2}. Since -\frac{1}{2} is negative, the inequality direction is changed.

$$x\leq \frac{1}{4}\left(-2\right)$$

Multiply \frac{1}{4} and -2 to get -\frac{1}{2}.

$$x\leq -\frac{1}{2}$$