The inequality

$$a + 2 \sqrt{a^{2} + b^{2}} + 3 \sqrt{1+a^{2}} \geq 2 + \sqrt{1+b^{2}}$$

is related to some computation regarding the Shapley value of a stochastic cooperative game. Though I'm not sure it holds, I suspect it is true for $a, b \in \mathbb{R}_{>0}$.

Do you know how to prove this inequality for $a, b \in \mathbb{R}_{>0}$? Or do you have any hints pointing towards a possible proof?

  • 3
    $\begingroup$ Seems straightforward: $2\sqrt {1+a^2}\ge2$ and $2\sqrt {a^2+b^2}+\sqrt{1+a^2}\ge 2b+1\ge\sqrt {1+b^2}$. $\endgroup$ – Andrés E. Caicedo Mar 5 '18 at 12:33
  • $\begingroup$ @AndrésE.Caicedo thank you! $\endgroup$ – Max Muller Mar 5 '18 at 12:38

Because $$a + 2 \sqrt{a^{2} + b^{2}} + 3 \sqrt{1+a^{2}} \geq 2b+3\geq b+3\geq2 + \sqrt{1+b^{2}},$$ where the last inequality it's $$b+1\geq\sqrt{1+b^2}$$ or $$2b\geq0,$$ which is obvious.

  • $\begingroup$ Thank you! Seems similar to Andrés E. Caicedo's answer. $\endgroup$ – Max Muller Mar 5 '18 at 12:40
  • $\begingroup$ You are welcome. Yes, but I wrote it before. ;) $\endgroup$ – Michael Rozenberg Mar 5 '18 at 12:49
  • $\begingroup$ That's true indeed! $\endgroup$ – Max Muller Mar 5 '18 at 14:58

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