Recognising always-non-negative sums

There is a family of inequalities that, once recognised, can be solved “by inspection”, without any calculation: those in which the left-hand side is a sum of quantities that can never be negative. A sum of radicals, or of absolute values, or of squares: each of these summands is by construction $\ge 0$ and hence so is their sum.

Property — Always-non-negative sums

Within their own existence conditions:

• $\sqrt{A(x)} + \sqrt{C(x)}$ is always $\ge 0$;
• $\sqrt{A(x)} + |C(x)|$ is always $\ge 0$;
• $|A(x)| + |C(x)|$ is always $\ge 0$;
• a sum of squares $\bigl(P(x)\bigr)^2 + \bigl(Q(x)\bigr)^2$ is always $\ge 0$;
• a distance (point–point or point–line) is always $\ge 0$.

Consequently, every inequality of the form $\underbrace{\sqrt{A(x)}+\sqrt{C(x)}}_{\text{o analoga}} \ge 0$ has as its solution the existence conditions alone; the same inequality with $<0$ (strict) is impossible (at most it may have an “isolated” solution at the points where all the summands are simultaneously zero).

Example — Three inequalities "by inspection"

$\sqrt{x} + \sqrt{x-1}\ge 0$. Existence conditions: $x\ge 0$ and $x\ge 1$, that is $x\ge 1$. The inequality is always true within the existence conditions, so the solution is $\boxed{\,x\ge 1\,}$.

$|x|+|x-1|<0$. It is a sum of two absolute values, always $\ge 0$. There is no $x$ for which it is strictly $<0$: no solution.

$\sqrt{x}+\sqrt{x^2-x}\le 0$. It is a sum of two radicals, $\ge 0$ within the existence conditions. It can equal zero only if both summands are zero simultaneously: $x=0$ and ($x=0 \vee x=1$). Sole acceptable value: $\boxed{\,x=0\,}$. If the inequality were strict ($<0$), there would be no solutions.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality · Absolute value
Methods: Non-negative sides shortcut
Skills: Reasoning by cases · Solving inequalities