VicKmath7 wrote: Solve the system in reals: $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2022$ and $x+y=\frac{2021}{\sqrt{2022}}$ Let $x=\cot 2u$ and $y=\cot 2v$ where $u,v\in\left(0,\frac{\pi}2\right)$ Let $z=\cot u$ and $t=\cot v$ System is $zt=2022$ and $z+t-\frac{z+t}{zt}=\frac{4042}{\sqrt{2022}}$ And so $zt=2022$ and $z+t=2\sqrt{2022}$ which immediately gives $z=t=\sqrt{2022}$ And so $\boxed{x=y=\frac{2021}{2\sqrt{2022}}}$

Simpler? (without these trigonometric substitutions)

VicKmath7 wrote: Simpler? (without these trigonometric substitutions) You are welcome. Glad to have helped you. I found this solution quite simple, according to me. Looking at the result, there is certainly an inequalities-property allowing to get $x=y$ But I have no skill in inequalities. And trig substitutions are simpler, for me, than inequalities properties (XM, UVWXYZ, Stolzivich and Co ) Simplicity seems quite relative

VicKmath7 wrote: Simpler? (without these trigonometric substitutions) Just because you have another solution to the problem that does not involve trig, does not make post no. 2 necessarily harder. Trig is a tool and can be used without any problems. As for the "simple" solution you are looking for, I think most people will find this simple.

VicKmath7 wrote: Solve the system in reals: $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2022$ and $x+y=\frac{2021}{\sqrt{2022}}$ Other solution (not trig-based but not simpler that my previous one, according to me, and still not inequality-based) Note that $x+y>0$ implies at least one of $x,y>0$ Let $f(x)=\ln(x+\sqrt{x^2+1})$ : $f(x)$ is continuous, odd, and concave over $\mathbb R^+$ Problem statement implies $f(x)+f(y)=f(\frac{x+y}2)$ If $x,y\ge 0$, this implies $x=y=\frac{x+y}2$ and so $\boxed{x=y=\frac{2021}{2\sqrt{2022}}}$, which indeed fits. If $x>0>y$ , this is $f(x)=f(\frac{x+y}2)+f(-y)$ and so $x+y=0$, impossible here.

VicKmath7 wrote: Solve the system in reals: $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2022$ and $x+y=\frac{2021}{\sqrt{2022}}$ Other solution (same than first, but with hidden trig, so maybe simpler ?) Let $u=x+\sqrt{x^2+1}$ so that $x=\frac 12(u-\frac 1u)$ Let $v=y+\sqrt{y^2+1}$ so that $y=\frac 12(v-\frac 1v)$ System is $uv=2022$ and $u+v-\frac{u+v}{uv}=\frac{4042}{\sqrt{2022}}$ And so $uv=2022$ and $u+v=2\sqrt{2022}$ which immediately gives $u=v=\sqrt{2022}$ And so $\boxed{x=y=\frac{2021}{2\sqrt{2022}}}$

That's the type of solution I was looking for, thanks .

pco your the last solution is interesting