Đáp án:

 điều phải chứng minh

Giải thích các bước giải:

\(\begin{array}{l}
1)Do:{x^2} \ge 0\forall x\\
 \to {x^2} + 5 > 0\forall x\\
4){x^2} + 2xy + {y^2} + 1\\
 = {\left( {x + y} \right)^2} + 1\\
Do:{\left( {x + y} \right)^2} \ge 0\forall x;y\\
 \to {\left( {x + y} \right)^2} + 1 > 0\\
7)3\left( {{x^2} + 2x + 1} \right)\\
 = 3{\left( {x + 1} \right)^2} \ge 0\forall x\\
2){x^2} + 2x + 1 = {\left( {x + 1} \right)^2} \ge 0\forall x\\
5){x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\\
8)5 + {x^2} + x - 2\\
 = {x^2} + x + 3\\
 = {x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{11}}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0\\
3){x^2} + 4x + 4\\
 = {\left( {x + 2} \right)^2} \ge 0\forall x\\
6){x^2} + 5x + 8\\
 = {x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} + \dfrac{7}{4}\\
 = {\left( {x + \dfrac{5}{2}} \right)^2} + \dfrac{7}{4}\\
Do:{\left( {x + \dfrac{5}{2}} \right)^2} \ge 0\forall x\\
 \to {\left( {x + \dfrac{5}{2}} \right)^2} + \dfrac{7}{4} > 0
\end{array}\)