Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?

It is not difficult to show that $x-\sqrt{\strut x^2+1}<0$ for all $x\in\mathbb{R}$. Then $$\left(x-\sqrt{\strut x^2+1}\right)\left(y-\sqrt{y^2+1}\right)>0.$$ We get $$xy+\sqrt{\strut x^2+1}\sqrt{y^2+1}>x\sqrt{y^2+1}+y\sqrt{\strut x^2+1}\tag{1}$$ From the other hand, since $(a-b)^2\geqslant 0$, then $$ab\leqslant\frac{a^2+b^2}{2}$$ for all $a,b\in\mathbb{R}$. Hence $$xy+\sqrt{\strut x^2+1}\sqrt{y^2+1}\leqslant\frac{x^2+y^2}{2}+\frac{x^2+1+y^2+1}{2}=x^2+y^2+1.$$ Joining this with (1) we get the desired inequality.
