Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} + 1 + \dfrac{1}{{{x^2} + 1}} \ge 2\sqrt {\left( {{x^2} + 1} \right).\dfrac{1}{{{x^2} + 1}}} \\
\Leftrightarrow {x^2} + 1 + \dfrac{1}{{{x^2} + 1}} \ge 2\\
\Leftrightarrow {x^2} + \dfrac{1}{{{x^2} + 1}} \ge 1,\forall x\\
\Rightarrow {x^2} + \dfrac{1}{{{x^2} + 1}} < 1\left( {VN} \right)\\
b)\sqrt {{x^2} - x + 1} + \dfrac{1}{{\sqrt {{x^2} - x + 1} }} \ge 2\sqrt {\sqrt {{x^2} - x + 1} .\dfrac{1}{{\sqrt {{x^2} - x + 1} }}} = 2\\
\Rightarrow \sqrt {{x^2} - x + 1} + \dfrac{1}{{\sqrt {{x^2} - x + 1} }} \ge 2,\forall x \in R\\
\Rightarrow \sqrt {{x^2} - x + 1} + \dfrac{1}{{\sqrt {{x^2} - x + 1} }} < 2\left( {VN} \right)\\
c)\sqrt {{x^2} + 1} + \sqrt {{x^4} - {x^2} + 1} - 2\sqrt[4]{{\left( {{x^6} + 1} \right)}}\\
= {\left( {\sqrt[4]{{{x^2} + 1}} - \sqrt[4]{{{x^4} - {x^2} + 1}}} \right)^2} \ge 0,\forall x\\
\Rightarrow \sqrt {{x^2} + 1} + \sqrt {{x^4} - {x^2} + 1} \ge 2\sqrt[4]{{\left( {{x^6} + 1} \right)}}\\
\Rightarrow \sqrt {{x^2} + 1} + \sqrt {{x^4} - {x^2} + 1} < 2\sqrt[4]{{\left( {{x^6} + 1} \right)}}\left( {VN} \right)
\end{array}\)
