Prove that $2\sqrt{x+1} \ge 2\sqrt{x} + 1/\sqrt{x+1}$ for $x \ge 1$

Question

How can I prove that the blue line is always less than or equal to the red line when X is greater than or equal to one?

This is a crucial algebraic step in a proof by mathematical induction. In order to complete my proof, I have to show that this is true. How do I do this?

Answer 1

We need to prove that $$2(x+1)\geq2\sqrt{x(x+1)}+1$$ or $$x+x+1\geq2\sqrt{x(x+1)},$$ which is true by AM-GM: $$\frac{x+(x+1)}{2}\geq\sqrt{x(x+1)}.$$