# $\sqrt{a^2+b^2} + \sqrt{b^2+4} \le \sqrt{(a+1)^2+b^2}+\sqrt{b^2+1}.$ [closed]

Let $$a,b \ge 2$$. I want to show that the following inequality is true

$$\sqrt{a^2+b^2} + \sqrt{b^2+4} \le \sqrt{(a+1)^2+b^2}+\sqrt{b^2+1}.$$

I have an intuition that the above inequality is true, but is there any elegant method to prove it without squaring both sides and expanding?

I tried by squaring both sides but lost in the calculation.

• You could use geometry. These are all diagonal distances with the LHS closer to a straight line
– Eric
Jul 30, 2021 at 12:12
• can you explain a bit more? @Eric Jul 30, 2021 at 12:13
• Please avoid math-only titles. These are discouraged for technical reasons - see Guidelines for good use of MathJax on question titles. Jul 30, 2021 at 12:32

Make a function $$f(x) = \sqrt{(x+1)^2+b^2} -\sqrt{x^2+b^2}$$
We need to prove $$f(a)\geq f(1)$$ which is true if $$f$$ is increasing on $$[1,\infty)$$. Calculate $$f'(x) ={x+1\over \sqrt{(x+1)^2+b^2}} - {x\over \sqrt{x^2+b^2}}$$ which can be easly seen that is positive and you are done.