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 ?
 A: 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.
A: Hint
For $x \ge 0$ and $y \ge 0$ use  
AM-GM
$$x^2+(y^2+1) \ge 2\sqrt{x^2(y^2+1)}=2x\sqrt{y^2+1}$$
$$(x^2+1)+y^2 \ge 2\sqrt{y^2(x^2+1)}=2y\sqrt{x^2+1}$$
Sum both and prove that the equality is not possible.
Also think about $x<0$ or $y<0$. That should be easy.
Can you finish?
A: \begin{align*}
x\sqrt{y^{2}+1}+y\sqrt{x^{2}+1}&=\left(x,y\right)\cdot\left(\sqrt{y^{2}+1},\sqrt{x^{2}+1}\right)\le\sqrt{(x^{2}+y^{2})(x^{2}+y^{2}+2)}\\
&\le\frac{2x^{2}+2y^{2}+2}{2}=x^{2}+y^{2}+1
\end{align*}
where I have used Cauchy-Schwarz followed by the arithmetic-geometric inequality. Note that equality cannot occur due to the equality case of the arithmetic geometric inequality.
A: Your inequality
is equivalent to the following one
$$
\frac12\left(x-\sqrt{y^2+1}\right)^2+\frac12\left(y-\sqrt{x^2+1}\right)^2>0
$$
which is clearly always true.
