\term{Universal generalization} is an inference rule so natural that
you've probably used it many times without a second thought.  Let $P$
be a predicate.  The rule says that in order to conclude:
%
\[
\forall \ x \in S \quad  P(x)
\]
%
it is sufficient to prove:
%
\[
x_0 \in S \implies P(x_0)
\]
%
where no additional assumptions are made about the value of $x_0$.
Less formally, if you can prove that a statement is true for some
arbitrary, unspecified $x_0 \in S$, then you can conclude that the
statement is true for all $x \in S$.  Use universal generalization to
prove that:
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\[
\forall \ x \in [-1,1] \quad 1 - x^2 \leq 1 - x_2 + x^4 - x^6
\]
%
Here [-1,1] denotes the set of all real numbers between -1 and 1,
including -1 and 1.  (If you prove this claim ``any old way'', you'll
probably discover in retrospect that you used universal
generalization.)
\end{problem}

\solution{Universal generalization says it is sufficent to prove:
%
\[
x_0 \in [-1,1] \implies
    1 - x_0^2 \leq 1 - x_0^2 + x_0^4 - x_0^6
\]
%
The standard recipe for proving an implication is to suppose that the
left side is true and then prove that the right side is also true,
subject to this supposition.  So suppose that $x_0 \in [-1,1]$.  Then
$x_0^2 \leq 1$, which means $1 - x_0^2$ is nonnegative.  For all $x_0
\in \mathbb{R}$, we have
%
\[
1 \leq 1 + x_0^4
\]
%
since $x_0^4$ is nonnegative.  Multiplying both sides by $1 - x_0^2$
does not reverse the inequality, so we have:
%
\begin{align*}
1 - x_0^2
    & \leq (1 + x_0^4) (1 - x_0^2) \\
    & = 1 - x_0^2 + x_0^4 - x_0^6
\end{align*}
}