<s>[INST] 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) $ 
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.
 [/INST] 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.
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