Bài làm:
Ta có: $\sqrt[]{3x^2+6x+12}$ = $\sqrt[]{3x^2+6x+3+9}$ = $\sqrt[]{(3x^2+6x+3)+9}$
= $\sqrt[]{3(x^2+2x+1)+9}$ = $\sqrt[]{3(x+1)^2+9}$
Vì $(x+1)^{2}$ $\geq$ 0 ⇒ 3$(x+1)^{2}$ + 9 $\geq$ 9
⇒ $\sqrt[]{3(x+1)^2+9}$ $\geq$ $\sqrt[]{9}$ = 3
hay $\sqrt[]{3x^2+6x+12}$ $\geq$ 3 (1)
Ta lại có: $\sqrt[]{5x^4-10x^2+9}$ = $\sqrt[]{5x^4-10x^2+5+4}$ = $\sqrt[]{5(x^4-2x^2+1)+4}$
= $\sqrt[]{5(x^2-1)^2+4}$
Vì ($x^{2}$ - 1)$^{2}$ $\geq$ 0 ⇒ 5($x^{2}$ - 1)$^{2}$ + 4 $\geq$ 4
⇒ $\sqrt[]{5(x^2-1)^2+4}$ $\geq$ $\sqrt[]{4}$ = 2
hay $\sqrt[]{5x^4-10x^2+9}$ $\geq$ 2 (2)
Từ (1) và (2) ⇒ $\sqrt[]{3x^2+6x+12}$ + $\sqrt[]{5x^4-10x^2+9}$ $\geq$ 2+3
⇒ $\sqrt[]{3x^2+6x+12}$ + $\sqrt[]{5x^4-10x^2+9}$ $\geq$ 5 (đpcm)
Dấu " = " xảy ra ⇔ $\left \{ {{x+1=0} \atop {x^2-1=0}} \right.$
⇔ \(\left[ \begin{array}{l}x=-1\\x^2=1\end{array} \right.\) ⇔ $\left \{ {{x=-1} \atop {x=±1}} \right.$ ⇔ x = -1
