@below ,Yes there was a miscalculation. Is your solution same, if not then post it.

I feel that the ans to this is $\frac{1}{4}$.

Solution: Note that the circle with coordinates $(x,x^2+k)$ has radius $x^2+k$. So, the circles with centers $(x,x^2+k)$ and with center $(y,y^2+k)$ meet at some point for any $x,y$. Triangle Inequality gives that this is equivalent to $(x^2+y^2+2k)^2 \geq (x^2-y^2)^2+(x-y)^2$. Since squares are always positive, this implies $x^2+y^2+2k \geq |x-y|$. The case is 'strongest' when $x,y$ have opposite signs. So we consider the case $|x|^2+|y|^2+2k \geq |x|+|y|$ which gives $2k \geq |x|-|x|^2+|y|-|y|^2$. Maxima is achieved at $x=y=\frac{1}{2}$ which gives $k \geq \frac{1}{4}$. Now we check whether the inequality $(x^2+y^2+\frac{1}{2})^2 \geq (x^2-y^2)^2+(x-y)^2$ is true or not. Expanding and cancelling gives that this is equivalent to $4x^2y^2+\frac{1}{4} \geq 2xy$ which is obvious by AM-GM. Thus $k=\frac{1}{4}$ is the smallest value.

Mathotsav wrote: Solution: Note that the circle with coordinates $(x,x^2+k)$ has radius $x^2+k$. So, the circles with centers $(x,x^2+k)$ and with center $(y,y^2+k)$ meet at some point for any $x,y$. Triangle Inequality gives that this is equivalent to $(x^2+y^2+2k)^2 \geq (x^2-y^2)^2+(x-y)^2$. Since squares are always positive, this implies $x^2+y^2+2k \geq |x-y|$. The case is 'strongest' when $x,y$ have opposite signs. So we consider the case $|x|^2+|y|^2+2k \geq |x|+|y|$ which gives $2k \geq |x|-|x|^2+|y|-|y|^2$. Maxima is achieved at $x=y=\frac{1}{2}$ which gives $k \geq \frac{1}{4}$. Now we check whether the inequality $(x^2+y^2+\frac{1}{2})^2 \geq (x^2-y^2)^2+(x-y)^2$ is true or not. Expanding and cancelling gives that this is equivalent to $4x^2y^2+\frac{1}{4} \geq 2xy$ which is obvious by AM-GM. Thus $k=\frac{1}{4}$ is the smallest value. Why is the triangle inequality $a+b\ge c $?

Maybe because degenerate triangle is possible?

Oh, I got it. Thanks

that paper was too hard then previous rmo paper

Red-Apple wrote: that paper was too hard then previous rmo paper [/quo No I think it was easier than previous

Suppose $\mathcal{C}_{1}$: a circle with center $(x,x^2+k)$ and radius $x^2+k$ (as it's tangent to $X$ axis) $\mathcal{C}_{2}$: a circle with center $(y, y^2+k)$ and radius $y^2+k$ (as it's tangent to $X$ axis) Now since they intersect at at least one point, from triangle inequality we have $(x^2+k+y^2+k)^2 \geqslant (x-y)^2+(x^2-y^2)^2 \implies x^2(4k-1+4y^2)+2xy+4ky^2-y^2+4k^2 \geqslant 0$ $\forall$ $x \in \mathbb{R}$(as choice of $\mathcal{C}_{1}$ , $\mathcal{C}_{2}$ is arbitrary). , So we have $\triangle_{x}=-16(-k^2+4k^3-2ky^2+8k^2y^2-y^4+4ky^4) \leqslant 0$, so we get $-16(4k-1)(k+y^2)^2 \leqslant 0 \implies 4k-1 \geqslant 0 \implies k \geqslant \frac{1}{4}$ , which gives minimum such real $k$ is $\frac{1}{4}$. Let's verify the case we found, we need to check if the triangle inequality holds for $k=\frac{1}{4}$ , we get $\left(x^2+y^2+\frac{1}{2}\right)^2 \geqslant (x^2-y^2)^2+(x-y)^2 \implies 4x^2y^2 +\frac{1}{4} \geqslant 2xy$ which is true by $\mathrm{AM-GM}$ inequality, hence $k=\boxed{\frac{1}{4}}$ is the required minimum value. $\blacksquare$

As no one decided to use Calc, here I go: We are basically looking for a $k$ such that the sum of radius of any two circle is greater than or equal to the distence between the centers, making the circles intersecting or tangent. This can be written as: $a^2+k+b^2+k \geq \sqrt{(a-b)^2+(a^2-b^2)^2}$ $\iff 2k \geq \sqrt{(a-b)^2+(a^2-b^2)^2}-a^2-b^2$ We are basically looking to minimize the RHS for positive reals $a,b$. We can do so by letting $a+b=C$ where $C$ is a constant. Then the RHS will be: $\sqrt{(C^2+1)(2a-C)^2}-2a^2+2Ca-C^2$ $= 2\sqrt{C^2+1}a+2Ca-2a^2-C\sqrt{C^2+1}-C^2$ We will minimize this using differentiation,: $2\sqrt{C^2+1}+2C-4a=0\\ \iff a= \frac{\sqrt{C^2+1}+C}{2}$ This means the RHS at it's minima is equal to: $C^2+1+C\sqrt{C^2+1}+2Ca-2a^2-C\sqrt{C^2+1}-C^2$ $= 1+C\sqrt{C^2+1}+C^2-\frac{1}{2}(2C^2+1+2C\sqrt{C^2+1})$ $= \frac{1}{2}$ Thus, $2k=\frac{1}{2}$, hence, $k= \frac{1}{4}$