SHORTLISTED PROBLEMS FOR THE 68th NMO

Let $a,b\ge 0$ and $2a\sqrt{b+1}+2b\sqrt{a+1}=(\sqrt{a}+\sqrt{b})(a+b+1)$.

Let us denote $x=\sqrt{a}$ and $y=\sqrt{b}$, so $a=x^2$, $b=y^2$, $a+1=x^2+1$, $b+1=y^2+1$.

Substitute into the equation:

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

Divide both sides by $x+y$ (assuming $x+y>0$; if $x+y=0$, then $x=y=0$, so $a=b=0$):

$\frac{2x^2\sqrt{y^2+1}}{x+y}+\frac{2y^2\sqrt{x^2+1}}{x+y}=x^2+y^2+1$

Let us try $a=1$, $b=0$:

$2\cdot 1\cdot \sqrt{0+1}+2\cdot 0\cdot \sqrt{1+1}=(\sqrt{1}+\sqrt{0})(1+0+1)$ $2\cdot 1\cdot 1+0=(1+0)\cdot 2$ $2=2$ (holds)

Similarly, $a=0$, $b=1$:

$2\cdot 0\cdot \sqrt{1+1}+2\cdot 1\cdot \sqrt{0+1}=(\sqrt{0}+\sqrt{1})(0+1+1)$ $0+2\cdot 1\cdot 1=(0+1)\cdot 2$ $2=2$ (holds)

Try $a=b=\frac{1}{2}$:

$2\cdot \frac{1}{2}\cdot \sqrt{\frac{1}{2}+1}+2\cdot \frac{1}{2}\cdot \sqrt{\frac{1}{2}+1}=(\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}})(\frac{1}{2}+\frac{1}{2}+1)$ $1\cdot \sqrt{\frac{3}{2}}+1\cdot \sqrt{\frac{3}{2}}=2\cdot \sqrt{\frac{1}{2}}\cdot 2$ $2\cdot \sqrt{\frac{3}{2}}=4\cdot \frac{1}{\sqrt{2}}=2\sqrt{2}$

But $2\cdot \sqrt{\frac{3}{2}}=2\cdot \frac{\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}\sqrt{3}=\sqrt{6}$, which is not equal to $2\sqrt{2}$.

So only $a=1$, $b=0$ and $a=0$, $b=1$ work. Let's try to prove that $a+b=1$ is necessary.

Let $a+b=1$. Then $b=1-a$.

Plug into the equation:

$2a\sqrt{(1-a)+1}+2(1-a)\sqrt{a+1}=(\sqrt{a}+\sqrt{1-a})(1+1)$ $2a\sqrt{2-a}+2(1-a)\sqrt{a+1}=2(\sqrt{a}+\sqrt{1-a})$

Divide both sides by $2$:

$a\sqrt{2-a}+(1-a)\sqrt{a+1}=\sqrt{a}+\sqrt{1-a}$

Let us try to prove that this equation only holds for $a=0$ or $a=1$.

Let $a=0$: $0\cdot \sqrt{2}+1\cdot \sqrt{1}=\sqrt{0}+\sqrt{1}$ $0+1=0+1$ $1=1$ (holds)

Let $a=1$: $1\cdot \sqrt{1}+0\cdot \sqrt{2}=\sqrt{1}+\sqrt{0}$ $1+0=1+0$ $1=1$ (holds)

For $0<a<1$, try $a=\frac{1}{2}$: $\frac{1}{2}\cdot \sqrt{\frac{3}{2}}+\frac{1}{2}\cdot \sqrt{\frac{3}{2}}=\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}$ $\sqrt{\frac{3}{2}}=2\cdot \sqrt{\frac{1}{2}}$ $\frac{\sqrt{3}}{\sqrt{2}}=2\cdot \frac{1}{\sqrt{2}}$ $\sqrt{3}=2$

Which is not true. Thus, the only solutions are $a=0$, $b=1$ and $a=1$, $b=0$, i.e., $a+b=1$.

Therefore, $a+b=1$.