Since $\sqrt{2x-1}$ is assumed to be real, then $x \geq \frac{1}{2}$. Let \[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A \] Squaring both sides we obtain \[ (1) 2x+2\sqrt{x^2-(2x-1)}=A^2 \] Since $x^2-(2x-1)=(x-1)^2$, we can write $(1)$ in the equivalent form \[ (2) x+|x-1|=\frac{A^2}{2} \] $(a)$ suppose $A=\sqrt{2}$; then $\frac{A^2}{2}=1$, and we look for solutions $x \geq \frac{1}{2}$ of \[ (3) x+|x-1|=1 \] For $\frac{1}{2} \leq x \leq 1, x-1 \leq 0$, so $|x-1|=x-1$ and (3) becomes \[ x+x-1=2x-1=1 \] This equation is satisfied only if $x=1$, which contradicts the assumption that $x>1$. We conclude that $A=\sqrt{2}$ if and only if $\frac{1}{2} \leq x \leq 1$. In view of this result, we can confine attention to the range $x>1$ in solving part $(b)$ and $(c)$. Then $x+|x-1|=2x-1$. $(b)$ If $A=1$, then $2x-1=\frac{A^2}{2}=\frac{1}{2}$, so $x=\frac{3}{4}$. Since this is not in the range $x>1$, there are no solutions for $A=1$. $(c)$ If $A=2$, then $2x-1=\frac{A^2}{2}=2$, so $x=\frac{3}{2}$

For IMO's problem, this is very easy, I think. For $a>b>0,$ use $\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b},\ \sqrt{a+b-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}$ On the condition that $2x-1\geq 0\Longleftrightarrow x\geq \frac{1}{2},$ remark that $(2x-1)+1=2x,\ (2x-1)\cdot 1=2x-1,$ we have $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{\frac{2x+\sqrt{2x-1}}{2}}+\sqrt{\frac{2x-\sqrt{2x-1}}{2}}$ $=\frac{\sqrt{2x-1}+1}{\sqrt{2}}+\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}=\begin{cases} \sqrt{2} & (\frac{1}{2}\leq x\leq 1) \\ \sqrt{2}\sqrt{2x-1} & (x\geq 1) \end{cases}$ Therefore sketching the graph of $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ on $x-A$ plane gives the desired answers immeadiately, that is to say, find the absissae of the intersection points for the graph of $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ and $A=the\ given\ constant.$ Therefore the desired answers are for $(a)\ \frac{1}{2}\leq x\leq 1,$ for $(b)$ there are no solutions and for $(c)\ x=\frac{3}{2}.$ kunny

I wish we could solve contemporary problems as easily as we do archaic ones.

So all we did is square both sides to make the left side a little friendlier?

I dont understand the different conclusions of the previous posts Using DPopov path (adding obviously $A\ge 0$), we get that the equation is equivalent to $x\ge\frac 12$ and $x+|x-1|=\frac {A^2}2$ Which is equivalent to : ($x> 1$ and $2x-1=\frac {A^2}2$) or ($1\ge x\ge \frac 12$ and $A^2=2$) $\iff$ ($x> 1$ and $x=\frac {A^2+2}4$) or ($1\ge x\ge \frac 12$ and $A^2=2$) And so : If $A>\sqrt 2$ : one solution $x=\frac {A^2+2}4$ If $A=\sqrt 2$ : infinitely many solutions : all $x\in[\frac 12,1]$ If $A<\sqrt 2$ : no solution.

DPopov wrote: For $\frac{1}{2} \leq x \leq 1, x-1 \leq 0$, so $|x-1|=x-1$ and (3) becomes \[ x+x-1=2x-1=1 \] Ummm....this is a mistake, isn't it?

Although I am not good at English reading,I think the describtion is well.The question is only solving three equations,it is not worth to solve all As.

DPopov wrote: For what real values of $x$ is \[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A \]given a) $A=\sqrt{2}$; b) $A=1$; c) $A=2$, where only non-negative real numbers are admitted for square roots? Multiply both sides by $\sqrt{2}$, we get $\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=A\sqrt{2}$ $\Leftrightarrow \sqrt{(2x-1)+2\sqrt{2x-1}+1}+\sqrt{(2x-1)-2\sqrt{2x-1}+1}=A\sqrt{2}$ $\Leftrightarrow \sqrt{(\sqrt{2x-1}+1)^2}+\sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}$ $\Leftrightarrow |\sqrt{2x-1}+1|+|\sqrt{2x-1}-1|=A\sqrt{2}$

Squaring, we have that: $$A^2 = 2x + 2\sqrt{x^2-2x+1} \implies A^2 = 2x+2|x-1|$$If $A=\sqrt{2}$, then, $2=2x+2|x-1| \implies 1-x=|x-1| \implies x \le 1$. Given that $\sqrt{2x-1} \in Z \implies x \ge \frac{1}{2}$, we have that $x \in \left[ \frac{1}{2},1\right]$. If $A=1$, then $1=2x+2|x-1| \implies x = \frac{3}{4}$, which we can check to be valid. If $A = 2$, then $4=2x+2|x-1| \implies 2-x=|x-1| \implies x = \frac{3}{2}$. However, if $x = \frac{3}{2}$, then $x-\sqrt{2x-1} < 0$, so there are no solutions.

For the inequality to hold, we must have: $x \geq \frac{1}{2}$ and $(x-1)^2 \geq 0$. note that if we square the identity we have the following \begin{align*} &\Rightarrow \left(\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}}\right)^2 \\ &\Rightarrow 2x + 2\left(\sqrt{x+\sqrt{2x-1}})(\sqrt{x-\sqrt{2x-1}}\right) \\ &\Rightarrow 2x+2\mid x-1 \mid. \end{align*}(The second follows by difference of squares.) From here it is easy to see: that $2x+\mid x-1 \mid = 2$ precisely when $\frac{1}{2} \leq x \leq 1$, $4-2x = 4 \Rightarrow x = \frac{3}{2}$ and there is no soltion for $1$

Why does it seem so much that early IMO problems are way easier than IMO problems nowadays?

I guess that at that time competition math wasn't so popular as it is nowadays

DPopov wrote: For what real values of $x$ is \[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A \]given a) $A=\sqrt{2}$; b) $A=1$; c) $A=2$, where only non-negative real numbers are admitted for square roots? Truly ancient problem

This one is from the first IMO 1959

We know that $2x-1\ge0$ so $x\ge\frac12$. We can square both sides of the equation $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A$ to get \begin{align*} x+\sqrt{2x-1}+2\sqrt{(x-\sqrt{2x-1})(x+\sqrt{2x-1})}+x-\sqrt{2x-1}&=A^2\\ 2x+2\sqrt{x^2-2x+1}&=A^2\\ 2x+2|x-1|&=A^2\\ x+|x-1|&=\frac{A^2}2 \end{align*} We can do casework on the equation $x+|x-1|=\frac{A^2}2$. We know that if $x\ge1$ then $x+|x-1|=x+x-1=2x-1$ so $2x-1=\frac{A^2}2$ and $x=\frac{A^2+2}4$. We also know that if $x\le1$ then $x+|x-1|=x+1-x=1$ so $1=\frac{A^2}2$ and $A=\sqrt2$. We can now solve for when $A=\sqrt2$, $A=1$, and $A=2$. (a) $A=\sqrt2$ We already figured out that when $A=\sqrt2$ that $x\le1$. We combine this with $x\ge\frac12$ to get $\frac12\le x\le1$. (b) $A=1$ We know that if $x\le1$ then $A=\sqrt2$ which is false. We know that if $x\ge1$ then $x=\frac{A^2+2}4=\frac34$ but $\frac34<1$ so this is also not a solution. Thus, there are no solutions. (c) $A=2$ We know that if $x\le1$ then $A=\sqrt2$ which is false. We know that if $x\ge1$ then $x=\frac{A^2+2}4=\frac32$. Thus, $x=\frac32$.

We have that $2x-1=0\iff x\geq\tfrac{1}{2}$ and $x+\sqrt{2x-1},x-\sqrt{2x-1}\geq0\iff x\geq0$. Henceforth assume that $x\geq\tfrac{1}{2}$. Since $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ and $A$ are both nonnegative, we may square both sides of the given equation to obtain an equivalent formulation: $$(x+\sqrt{2x-1})+(x-\sqrt{2x-1})+2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=(x+x)+2\sqrt{x^2-2x+1}=2x+2|x-1|=A^2,$$or, equivalently, $x+|x-1|=\tfrac{A^2}{2}$. For (a) $A=\sqrt{2}$, when $x\leq1$, we have $x+(1-x)=1=\tfrac{(\sqrt{2})^2}{2},$ always true; and when $x\geq1$, we have $x+(x-1)=1\iff x=1$. Hence, the solutions to (a) are given by $\boxed{x\in[\tfrac{1}{2},1]}$. For (b) and (c), we have $x+(1-x)=1=\tfrac{A^2}{2}$ when $x\leq1$, never true, so it suffices to consider $x\geq1$. For (b), $$x+(x-1)=\frac{1^2}{2}=\frac{1}{2}\iff x=\frac{3}{4}\not\geq1,$$so $\boxed{\text{no solutions}}$; and for (c), $$x+(x-1)=\frac{2^2}{2}=2\iff\boxed{x=\frac{3}{2}}\geq1.$$$\blacksquare$

Note that $x\geq 1/2.$ Squaring it we get $2x+2|x-1|=A^2.$ Easily -- a) $\implies x\in [1/2,1]$ b) $\implies \not\exists x$ b) $\implies x=3/2$

We can square this expressed to get $2x+|x-1|=A^2/2.$ we can now substitute values of a and solve

$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A$ $x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=A^2$ $2x+2\sqrt{x^2-(2x-1)}=A^2$ $2\sqrt{x^2-(2x-1)}=A^2-2x$ $4x^2-4(2x-1)=A^4+4x^2-4A^2x$ $x(4A^2-8)+(4-A^4)=0$ Plugging $A=\sqrt{2}$ results in all solutions, but for this to work, $2x-1 \ge 0, x \ge \frac{1}{2}$. Also, $x-\sqrt{2x-1} \ge 0, x \ge \sqrt{2x-1}, x^2 \ge 2x-1, x^2-2x+1 \ge 0, (x-1)^2 \ge 0$, which is always true. Also, $A^2-2x \ge 0, 2-2x \ge 0, -2x \ge -2, x \ge 1$ So $\boxed{\frac{1}{2} \le x \le 1}$ Plugging in $A=1$ results in $x=\frac{3}{4}$, but $A^2-2x = -\frac{1}{2} < 0$, so $\boxed{\text{No Solution}}$ Plugging in $A=2$ results in $x=\frac{3}{2}$, which satisfies $A^2-2x \ge 0$ and $2x-1 \ge 0$, so $\boxed{x=\frac{3}{2}}$

$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$, where $c^2=a^2-b$ Plug it in and trivializes