Një pabarazi si çdo tjetër // An Inequality like any other

(Albanian)          

                     A photo of a solution from Facebook looks like this. She is trying to  solve a problem proposed by Bogdan ENESCU, from an unspecified source. The same man raised the problem a year ago.

        Statement:

                           Let $x,y \in \mathbb{R}.$ Prove that

1. If $(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1$, then $x+y=0;$

2. If $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1$, then $x+y=0.$

Many solutions, the most elegant using the hyperbolic trigonometric function substitution, appear in the comments. Here we only point out an algebraic inequality that solves point 2. of the problem.

  

          LEMMA The inequality 

$$\sqrt{1+x^2} \cdot \sqrt{1+y^2}\geqslant |1-x\cdot y| \tag{1}$$

           holds, for any $x,y \in \mathbb{R}$; the "$=$" sign occurs if and only if $x+y=0.$

          Proof of the lemma

$(x+y)^2 \geqslant 0\;\Leftrightarrow \;x^2+2xy+y^2 \geqslant 0\;\Leftrightarrow \;1+x^2+y^2+x^2 \cdot y^2 \geqslant 1-2xy+x^2\cdot y^2\;$

$\Leftrightarrow \;(1+x^2)(1+y^2)\geqslant (1-xy)^2\;\Leftrightarrow \;(1)$.

$\square$

        Now $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1\;\Rightarrow$

$$\Rightarrow\;xy+y\sqrt{1+y^2}+x\sqrt{1+x^2}+\sqrt{1+x^2}\sqrt{1+y^2}=1\;\Rightarrow$$

$$\Rightarrow\;y\sqrt{1+y^2}+x\sqrt{1+x^2}=1-xy-\sqrt{1+x^2}\sqrt{1+y^2}\;\overset{squaring}{\Rightarrow}$$

$$\Rightarrow\;y^2(1+y^2)+x^2(1+x^2)+2xy\sqrt{1+x^2}\sqrt{1+y^2}=$$

$$=1+x^2y^2+(1+x^2)(1+y^2)-2xy-2\sqrt{1+x^2}\sqrt{1+y^2}+2xy\sqrt{1+x^2}\sqrt{1+y^2}\;\Rightarrow$$

$$\Rightarrow\;x^4+y^4-2x^2y^2=2-2xy-2\sqrt{1+x^2}\sqrt{1+y^2}\;\Rightarrow$$

$$\Rightarrow\;1-xy-\sqrt{1+x^2}\sqrt{1+y^2}=\frac{(x^2-y^2)^2}{2}.\tag{2}$$

But, from the Lemma we have

$1-xy\leqslant |1-xy|\leqslant \sqrt{1+x^2}\sqrt{1+y^2}$. So $1-xy-\sqrt{1+x^2}\sqrt{1+y^2}\leqslant 0$

and then from (2) it results $\frac{(x^2-y^2)^2}{2}\leqslant 0$. Hence $x^2=y^2$.

If $x=y$, then 2. is written $(x+\sqrt{1+x^2})^2=1$ which is successively equivalent to $x^2+(1+x^2)+2x\sqrt{1+x^2}=1\;\Leftrightarrow\;x^2+x\sqrt{1+x^2}=0$, hence $x=0=y.$

If $x=-y$, then $(x+\sqrt{1+y^2})(x+\sqrt{1+x^2})=(x+\sqrt{1+x^2})(-x+\sqrt{1+x^2})=(1+x^2)-x^2=1.$ In both cases 2. is fulfilled.

         Tanks to Doru Stoica for this solution.