Prove or disprove: $7 \lt \sqrt{3} + \sqrt{27}$

Question

In an admission test to enroll in a Earth's Science Bachelor Degree course there is this question:

Sort in increasing order $7$, $\sqrt{47}$ and $\sqrt{3} + \sqrt{27}$.

Now, I know that $7=\sqrt{49}$ and $\sqrt{x}$ is an increasing function and so from $x_1\lt x_2$ it follows that $\sqrt{x_1} \lt \sqrt{x_2}$; hence $\sqrt{47}\lt \sqrt{49}=7$.

But is it true that $7 \lt \sqrt{3} + \sqrt{27}$? How can I prove or disprove it?

Taylor approximation?

Some paper and pencil algorithm to compute an approximation of the root? (I learnt it at age of 12 but immediately forgot it).

Answer 1

Assuming $$ 7< √3 + √27 $$

Square both sides,

$$ 49<30+2√81 $$

Are we in a good position to continue OP? You will see that the above equation is coming false which disproves $ 7< √3 + √27 $.

Answer 2

$\sqrt{27}=3\sqrt{3};$

RHS: $\sqrt{3}+3\sqrt{3}=4\sqrt{3}. $

Square:

$7^2 <(?)16 \cdot 3.$

Since $f(x) : =\sqrt{x}$ is stricly increasing, it follows that the inequality does not hold.