Squared length triangle inequalities

Note that if $0<x<1,$ then $x>x^2,$ and if $x\ge 1,$ then $x^2 \ge x.$

Case 1 : $0<x < \frac{1}{2}.$

$\frac{1}{2}$ is the length of the longest side. By the triangle inequality ,

$x^2 + x > \frac{1}{2}.$

Solving this inequality gives

$x<\frac{-1-\sqrt{3}}{2} \cup x>\frac{-1+\sqrt{3}}{2}.$

Subject to the conditions of this case, the interval is $\left(\frac{-1+\sqrt{3}}{2},\frac{1}{2}\right).$

Case 2 : $\frac{1}{2} \le x<1.$

$x$ is the length of the longest side. By the triangle inequality,

$x^2+\frac{1}{2} > x.$

The solution to this inequality is $x\in\mathbb{R}.$ Subject to the conditions of this case, the interval is $\left[\frac{1}{2},1\right).$

Case 3 : $x \ge 1.$

$x^2$ is the length of the longest side. By the triangle inequality,

$x+\frac{1}{2} > x^2.$

Solving this inequality gives

$\frac{1-\sqrt{3}}{2} < x < \frac{1+\sqrt{3}}{2}.$

Subject to the conditions of this case, the interval is $\left[1,\frac{1+\sqrt{3}}{2}\right).$

Taking into account all cases, the interval of values that $x$ can take is:

$\left(\frac{-1+\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}\right).$

By triangle inequality,

$x^{2} + x > 1/2$

$\Rightarrow 2x^{2} +2x - 1 > 0$

which can be solved by wavy curve, to get $\frac{-1 \pm \sqrt{3}}{2}$ as critical points, but we can ignore the negative value i.e. $\frac{-1 - \sqrt{3}}{2}$ .

Now, the region ( $\frac{-1 + \sqrt{3}}{2},\infty$ ) gives positive values.

Similarly,

$x + 1/2 > x^{2}$

$\Rightarrow 2x^{2} -2x - 1 < 0$

By wavy curve (ignoring the negative values) the region ( $0,\frac{1 + \sqrt{3}}{2}$ ) gives negative values.

Also,

$x^{2} + 1/2 > x$

$\Rightarrow 2x^{2} -2x +1 > 0$

$\Rightarrow x^{2} + (x -1)^{2} > 0$

Which is always true since x is positive. ( Equality doesn't hold since $x > 0$ )

We require the region common in both i.e. ( $\frac{-1 + \sqrt{3}}{2},\frac{1 + \sqrt{3}}{2}$ )