Prove $\sqrt{n+1} < \sqrt{n} + 1 $ Prove $\sqrt{n+1} &lt; \sqrt{n} + 1 $ for all $n \ge 1$. 
I have proven the base step for $n = 1$. $\sqrt{2}$ is less than $2$. The inductive hypothesis is $\sqrt{n+1} &lt; \sqrt{n} + 1$. From here, I am unsure of how to proceed. How can I show that the inequality $\sqrt{(k+1)+ 1} &lt; \sqrt{k+1} + 1$? After simplifying the expression $\sqrt{(k+1)+1}$, it becomes $\sqrt{k+2}$. Perhaps I can use this somehow to come up with an expression? The right hand side does not have a radical over the 1, so I do not think it is a duplicate question that has an answer in the suggested question.
A geometric proof:

Consider two segments of lengths $1$ and $\sqrt{n}$. If you attach these segments by their endpoints at an angle of $90^\circ$ (so they become two legs of a right triangle), then by Pythagorean theorem the hypotenuse of this right triangle is $\sqrt{n+1}$.

Now increase the angle between the two "leg" segments. By the law of cosines, the distance between the "loose" endpoints also increases, until the angle becomes $180^\circ$. But now at $180^\circ$, the distance between the "loose" endpoints is just the sum of the two segments $\sqrt{n}+1$; therefore
$$\sqrt{n+1} &lt; \sqrt{n}+1.
$$
Q.E.D.
To prove by induction:
Consider n=1, $\sqrt {1+1} \lt \sqrt{1} + 1$
$$\sqrt{2} \lt 2$$
Assume n=k true:
$$\sqrt{k+1}\lt \sqrt{k} + 1$$
Consider n=k+1:
If true, this would mean that:
$$\sqrt{k+2}\lt\sqrt{k+1} + 1$$
We know that $\sqrt{k+1}\lt \sqrt{k} + 1$, so:
$$\sqrt{k+1} + 1 \lt \sqrt{k} + 2$$
Hence:
$$\sqrt{k+2}\lt\sqrt{k+1} + 1\lt\sqrt{k} + 2$$
Squaring both sides:
$$k+2\lt k + 4 + 4\sqrt{k}$$
This is indeed true, so n=1 true, and n=k true implies that n=k+1 true hence true for all natural numbers (integers greater than or equal to 1).