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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.

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  • $\begingroup$ You could use geometry. These are all diagonal distances with the LHS closer to a straight line $\endgroup$
    – Eric
    Jul 30, 2021 at 12:12
  • $\begingroup$ can you explain a bit more? @Eric $\endgroup$
    – User8976
    Jul 30, 2021 at 12:13
  • $\begingroup$ Please avoid math-only titles. These are discouraged for technical reasons - see Guidelines for good use of MathJax on question titles. $\endgroup$
    – soupless
    Jul 30, 2021 at 12:32

2 Answers 2

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The path closer to the diagonal is your LHS. It’s not a rigorous proof, but it provides pretty strong intuition.

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  • $\begingroup$ This can be made rigorous if you apply the triangle inequality twice or three times :) [+1] $\endgroup$
    – Dr. Mathva
    Aug 15, 2021 at 9:22
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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.

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