Since $\sqrt{x^4+7x^3+x^2+7x}+3\sqrt{3x}\ge10x-x^2$ is true for $x\ge 10$, the proof can continue for the interval $x<10$.
Squaring both sides gives:
$2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}+x^4+7x^3+x^2+34x \ge x^4-20x^3+100x^2$$\dfrac{2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}}{\sqrt{x}}\ge -27x^2+99x-34$
Now we have 2 cases, one of which the RHS is positive, and one of which where it is negative.
Case I:
$0<x<\dfrac{-\sqrt{681}+33}{18}$ or $\dfrac{\sqrt{681}+33}{18}<x<10$ These give RHS a negative, so the relation holds.
Case II:
$\dfrac{-\sqrt{681}+33}{18}<x<\dfrac{\sqrt{681}+33}{18}$
RHS is positive in this interval, so squaring both sides again we have
$\left(\dfrac{2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}}{\sqrt{x}}\right) ^2 \ge (-27x^2+99x-34)^2$
$-729x^4+5454x^3-10881x^2+6840x-400\ge 0$
$-(3x-4)^2(81x^2-390x+25) \ge 0$ and this inequality holds true in the interval $\dfrac{-\sqrt{681}+33}{18}<x<\dfrac{\sqrt{681}+33}{18}$ and therefore, we can conclude that $\sqrt{x^4+7x^3+x^2+7x}+3\sqrt{3x}\ge10x-x^2$