$\begin{array}{l}
co\;bat\;dang\;thuc\;minkopski:\sqrt {{a^2} + {b^2}} + \sqrt {{c^2} + {d^2}} \ge \sqrt {{{\left( {a + c} \right)}^2} + {{\left( {b + d} \right)}^2}} (1)\\
cm:\\
(1) \leftrightarrow {a^2} + {b^2} + {c^2} + {d^2} + 2\sqrt {\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)} \ge {a^2} + {b^2} + {c^2} + {d^2} + 2ac + 2bd\;(binh\;phuong\;2\;ve)\\
\leftrightarrow \sqrt {\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)} \ge ac + bd\;(day\;chinh\;la\;bat\;dang\;thuc\;bunhiacopski\;dung\;\forall a,b,c,d \in R)\\
dau = \leftrightarrow \frac{a}{c} = \frac{b}{d}\\
\to \sqrt {{a^2} - ab + {b^2}} + \sqrt {{b^2} - bc + {c^2}} = \sqrt {{{\left( {\frac{a}{2} - b} \right)}^2} + \frac{{3{a^2}}}{4}} + \sqrt {{{\left( {b - \frac{c}{2}} \right)}^2} + \frac{{3{c^2}}}{4}} \\
\ge \sqrt {{{\left( {\frac{a}{2} - b + b - \frac{c}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 a}}{2} + \frac{{\sqrt 3 c}}{2}} \right)}^2}} = \sqrt {{a^2} + ac + {c^2}} \\
dau = \leftrightarrow \frac{{\frac{a}{2} - b}}{{b - \frac{c}{2}}} = \frac{a}{c}
\end{array}$
$\eqalign{
& do\;n \in N \to {x^{2n}},{x^{2n - 2}},....1 \ge 0 \cr
& \cos i: \cr
& {{{x^{2n}}} \over 2} + {{{x^{2n - 2}}} \over 2} \ge 2\sqrt {{{{x^{4n - 2}}} \over 4}} = \left| {{x^{2n - 1}}} \right| \cr
& ma:\left| {{x^{2n - 1}}} \right| \ge \mp {x^{2n - 1}}(tinh\;chat\;cua\;tri\;tuyet\;doi) \to {{{x^{2n}}} \over 2} + {{{x^{2n - 2}}} \over 2} \pm {x^{2n - 1}} \ge 0 \cr
& \to f(x) = {x^{2n}} \pm {x^{2n - 1}} + {x^{2n - 2}} \pm {x^{2n - 3}} + ... + 1 \cr
& = {{{x^{2n}}} \over 2} + \left( {{{{x^{2n}}} \over 2} \pm {x^{2n - 1}} + {{{x^{2n - 2}}} \over 2}} \right) + \left( {{{{x^{2n - 2}}} \over 2} \pm {x^{2n - 3}} + {{{x^{2n - 4}}} \over 2}} \right) + ... + {1 \over 2} \cr
& \ge {{{x^{2n}}} \over 2} + {1 \over 2} \ge {1 \over 2} \cr
& dau = \cr
& \leftrightarrow \left\{ \matrix{
x = 0 \cr
{{{x^2}} \over 2} = {1 \over 2} \cr} \right. \leftrightarrow \overline \exists x \cr
& \to f(x) > {1 \over 2} \cr} $