shohvanilu wrote: Find the minimum value of $|x-y|+\sqrt{(x+2)^2+(y-4)^4}$ You should change your source of olympiad training problems (or stop just inventing crazy problems) Replacing $x\to x-2$ and $y\to a+4$, we can define $f(x)=|x-a-6|+\sqrt{x^2+a^4}$ If $x\ge a+6$, we get $f'(x)=\frac{x+\sqrt{x^2+a^4}}{\sqrt{x^2+a^4}}\ge 0$ and so minimum is reached when $x=a+6$ If $x\le a+6$, we get $f'(x)=\frac{x-\sqrt{x^2+a^4}}{\sqrt{x^2+a^4}}\le 0$ and so minimum is reached when $x=a+6$ $f(a+6)=\sqrt{(a+6)^2+a^4}$ and we are looking for $a$ such that this value is minimum So we are looking for the minimum of $g(x)=x^4+(x+6)^2=x^4+x^2+12x+36$ $g'(x)=4x^3+2x+12$ $g"(x)=12x^2+2>0$ So $g(x)$ is decreasing on $(-\infty,r]$ and increasing on $[r,+\infty)$ and the requested value is $\sqrt{g(r)}$ where $r$ is the unique real root of $x^3+\frac 12x+3=0$ This is a standard basic cubic in Cardano's form and we immediately get $r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}}$ $-\sqrt[3]{\frac{\sqrt{2922}+54}{36}}$ Hence the answer : Minimum is $\sqrt{r^4+r^2+12r+36}$ where $r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}}$ $-\sqrt[3]{\frac{\sqrt{2922}+54}{36}}$ Numeric application gives : $r\sim -1.32695628567897...$ $\min_{x,y\in\mathbb R}(|x-y|+\sqrt{(x+2)^2+(y-4)^4}$ $\boxed{\sim 4.99377611842389...}$

I hope he meant $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$ (the power $4$ was just a typo); then at least it has a nice geometric meaning and a nice, tame solution.

mavropnevma wrote: I hope he meant $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$ (the power $4$ was just a typo); then at least it has a nice geometric meaning and a nice, tame solution. Maybe it was $|x-y|+\sqrt{(x+2)^0+(y-4)^0}$, which seems easy too.

is there a solution for $|x-y|+\sqrt{(x+2)^2+(y-4)^2}$

You can fix $|x-y|=r>0$. Then for each $r$, let $f(r)$ be the minimum distance (in $r$) of any line in the curve $|x-y|=r$ to the point $(-2,4)$. Then, minimize $r+f(r)$.

Mavropnevma you right it's original version of this problem.

Sardor wrote: Mavropnevma you right it's original version of this problem. $|x-y|+\sqrt{(x+2)^2+(y-4)^2}\ge|x-y|+\frac{\sqrt{2}}{2}|(x+2)-(y-4)|$ $=|x-y|+|3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y)|\ge |x-y|+3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y)$ $\ge (1-\frac{\sqrt{2}}{2})|x-y|+3\sqrt{2}\ge 3\sqrt{2}$

CAn we substitute x by cost-2 and y by sint +4?