Which is bigger and how to check?

Question

I have $\dfrac{20\sqrt3-23}3$ and $\dfrac{\sqrt6+12}3$ but I don't know how to check which is bigger and which is smaller. Can someone help me?

Answer 1

The question is $\dfrac{20\sqrt3-23}3?\dfrac{\sqrt6+12}3$

Multiply both sides by $3$, then add $23$:
$20\sqrt3$ $?$ $\sqrt6+35$

Now $\sqrt3<1.8$, as $3<3.24=1.8^2$ and $\sqrt6>1$
So $20\sqrt3<36=1+35<\sqrt6+35$

$\therefore\dfrac{20\sqrt3-23}3<\dfrac{\sqrt6+12}3$

Answer 2

Once you've guessed which one is bigger (either by luck or by checking rational approximations), the job is simply finding tight enough rational bounds for the irrational quantities. In particular, note that $\sqrt{3}<\frac{9}{5}$ and $\sqrt{6}>2$. Then: $$\frac{20\sqrt{3}-23}{3}<\frac{36-23}{3}=\frac{13}{3}$$ While $$\frac{\sqrt{6}+12}{3}>\frac{2+12}{3}=\frac{14}{3}$$ So $\frac{\sqrt{6}+12}{3}$ is bigger.