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Can anybody help me prove this inequality ?

4 Answers 4

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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.

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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?

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    Haha, I need more practice with AM-GM. – 2017-01-11
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    @SimpleArt: For all three solutions you can see here. – 2017-01-11
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\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.

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    Very nice solution. – 2017-01-11
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    @szw1710 Thank you very much. – 2017-01-11
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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.