Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $

by c-2785   Last Updated February 13, 2018 13:20 PM

For positive real $a, b$, $$ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right).$$

I know the inequality $ (a+b)^2 \le 2(a^2 +b^2 ) $, but this is not useful for me. Then I think the function $$f(x) = \frac{ x^2 } {\sqrt{x^2 +1 }},$$ and the inequality is $$f(a+b) <2 ( f(a) + f(b)).$$

But I meet a wall and I don't know if this inequality is true. I want some hints. Thank you.



Answers 2


The inequality you know gives you

$$\frac{(a + b)^2}{\sqrt{(a + b)^2 + 1}} \le 2\left(\frac {a^2}{\sqrt{(a + b)^2 + 1}} + \frac{b^2}{\sqrt{(a + b)^2 + 1}}\right).$$

Now just remember that $(a + b)^2 > a^2$ as well, so

$$\frac{1}{\sqrt{(a + b)^2 + 1}} < \frac{1}{\sqrt{a^2 + 1}}$$

and so on.

user296602
user296602
February 13, 2018 12:36 PM

By C-S we obtain: $$2\left(\frac {a^2}{\sqrt{a^2 + 1}} + \frac{b^2}{\sqrt{b^2 + 1}}\right)\geq\frac{2(a+b)^2}{\sqrt{a^2+1}+\sqrt{b^2+1}}\geq$$ $$\geq\frac{2(a+b)^2}{\sqrt{(1^2+1^2)(a^2+1+b^2+1)}}=\frac{(a+b)^2}{\sqrt{\frac{a^2+b^2}{2}+1}}\geq\frac{(a+b)^2}{\sqrt{(a+b)^2+1}}.$$

Michael Rozenberg
Michael Rozenberg
February 13, 2018 13:12 PM