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