The first equation becomes $(ab+\sqrt{a^2+b}\sqrt{a+b^2})^2=(-\sqrt{ab+1})^2$. This rearranges to $a^3+2a^2b^2+b^3+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$. Now let $x=b\sqrt{a^2+b}+a\sqrt{b^2+a}$; then we get $x^2=a^3+2a^2b^3+b^2+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$, so $x=\pm 1$. Now suppose $x=-1$. Note that $ab\leq 0$, since $\sqrt{a^2+b}\sqrt{a+b^2}+\sqrt{ab+1}\geq 0$. Then one of the variables is $\leq 0$, one is $\geq 0$. WLOG let $a\leq 0, b\geq 0$. Then by $x=-1$ we get $(a\sqrt{b^2+a})^2=(-1-b\sqrt{a^2+b})^2$, which rearranges to $a^3-b^3=1+2b\sqrt{a^2+b}$. But since $a^3-b^3\leq 0$ and $\geq 0$, this is impossible, so $\boxed{x=1}$. Side note, is a construction needed for this problem? The phrasing made it seem like you didn't, and I couldn't actually find one in contest (I think the exact phrasing is something like "Suppose real numbers $a, b$ satisfy...find, with prove, the value of...")

Nice I got the answer $1$ as well But I kinda guessed. I cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds Assume that $b>0,a<0$ and then I got $b^3=\frac{\sqrt{5}-1}{2}$, the desired value is $\sqrt{b^6+b^3}=1$

I somehow provided a wrong construction for both $x=1$ and also somehow thought $x=-1$ worked because I found a wrong construction.

Bluesoul wrote: Nice I got the answer $1$ as well But I kinda guessed. I cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds Assume that $b>0,a<0$ and then I got $b^3=\frac{\sqrt{5}-1}{2}$, the desired value is $\sqrt{b^6+b^3}=1$ This seems false....... and wildly unnecessary

Assume that real numbers $a$ and $b$ satisfy $$ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0.$$Find, with proof, the value of $$b\sqrt{a^2+b}+a\sqrt{b^2+a}.$$

Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $$b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$$ Good question. 2022CJMO

sqing wrote: Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $$b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$$

sqing wrote: sqing wrote: Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $$b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$$ Where did you find this Chinese solution? Or you wrote this solution?

Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a+b^2)(b+a^2)}=5.$ Prove that $$2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4 $$

My solution, I wrote up in last 5 minutes so there were a lot of skipping. I wouldn't be surprised if this got a 0 or 1, though i may be slightly disappointed. P.S. : Is there a collection made for this year's CMO yet?

Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a^2+b)(b^2+a)}=5.$ Prove that $$2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4$$

sqing wrote: Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a^2+b)(b^2+a)}=5.$ Prove that $$2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4$$

Notice that \begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*}Now obviously $ab<0$, suppose $a>0>b$, then $a^3>b^3$ which implies $$b^2(a^2+b)<a^2(b^2+a)$$Hence $|b\sqrt{a^2+b}|<|a\sqrt{b^2+a}|$ and so $$b\sqrt{a^2+b}+a\sqrt{b^2+a}>0$$which implies that the value is $1$.

mathaddiction wrote: Notice that \begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*} A large part of the problem is actually proving whether the value is $1$ or $-1$

MortemEtInteritum wrote: mathaddiction wrote: Notice that \begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*} A large part of the problem is actually proving whether the value is $1$ or $-1$ Indeed! Thanks very much for pointing out, just edited.

Does this work??? Rewrite the given condition as $$ab+\sqrt{a^2+b} \cdot \sqrt{b^2+a} = -\sqrt{ab+1}$$$$a^3 + a b + 2 a^2 b^2 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{a + b^2}=ab+1$$$$a^3+ 2 a^2 b^2 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{a + b^2}=1$$$$(a\sqrt{b^2+a}+b\sqrt{a^2+b})^2=1$$$$a\sqrt{b^2+a}+b\sqrt{a^2+b}= \pm 1$$ We prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $$a\sqrt{b^2+a}+b\sqrt{a^2+b}=a\sqrt{c^2+a}-c\sqrt{a^2-c}.$$Note that we must have $a^2 \geq c$ or $a \geq \sqrt{c}.$ $$a\sqrt{c^2+a}-c\sqrt{a^2-c} \geq \sqrt{c^3+c\sqrt{c}}\geq 0.$$

Any other approach to solve this problem? :')

From the first equation, \[ a^2b^2 + ab + 1 + 2ab\sqrt{ab + 1} = (a^2 + b)(b^2 + a) = a^2b^2 + b^3 + a^3 + ab,\]\[ 1 + 2ab\sqrt{ab + 1} = a^3 +b^3.\]Let $k$ be the desired quantity. Then, \[ k^2 = 2a^2b^2 + a^3 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{b^2 + a},\]\[ k^2 = 2a^2b^2 + 2ab\sqrt{ab + 1} + 1 + 2ab(-ab - \sqrt{ab + 1}) = 1.\]Hence, $k = \pm 1$. We claim $k = -1$ is impossible, so assume for contradiction $k = -1$. Note that $ab \le 0$, so $a\le 0\le b$ or $b\le 0\le a$; WLOG let it be the second one. We have $b\sqrt{a^2 + b} + a\sqrt{b^2 + a} = -1$, and subtracting $a\sqrt{b^2 + a}$ and squaring gives \[ b^2(a^2 + b) = 1 + a^2(b^2 + a) + 2a\sqrt{b^2 + a}\]\[ \implies b^3 = 1 + a^3 + 2a\sqrt{b^2 + a}\implies b^3 - a^3 = 1 + 2a\sqrt{b^2 +a}.\]But this is impossible because $b^3 - a^3\le 0$ and $1 + 2a\sqrt{b^2 +a}\ge 1$. Hence, $k = +1$.

samrocksnature wrote: We prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $$a\sqrt{b^2+a}+b\sqrt{a^2+b}=a\sqrt{c^2+a}-c\sqrt{a^2-c}.$$Note that we must have $a^2 \geq c$ or $a \geq \sqrt{c}.$ $$a\sqrt{c^2+a}-c\sqrt{a^2-c} \geq \sqrt{c^3+c\sqrt{c}}\geq 0.$$ I may be misunderstanding some part, but the first part of the last inequality isn't necessarily true, right?. Consider a=3 and c=4, where $3\sqrt{19}-4\sqrt{5} \approx 4.13 < \sqrt{72} \approx 8.49$.