For any reals $x, y$, show the following inequality: $$\sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} \le \sqrt{(x-2)^2 + (y-6)^2} + \sqrt{(x-5)^2 + (y-6)^2} + 20$$ (Proposed by Bogdan Rublov)
Problem
Source: Kyiv City MO 2022 Round 1, Problem 9.2
Tags: algebra, inequalities
24.01.2022 04:14
MS_Kekas wrote: For any reals $x, y$, show the following inequality: $$\sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} \le \sqrt{(x-2)^2 + (y-6)^2} + \sqrt{(x-5)^2 + (y-6)^2} + 20$$ (Proposed by Bogdan Rublov) Hardest algebra ever? is it?
24.01.2022 04:19
@above there's something called sarcasm and hyperbolism in language.
25.01.2022 19:39
Apparently, equality cannot be reached ($(x,y)=(5,6)\Rightarrow 22.04=12.04+10=\sqrt{145}+10<3+20=23$)
26.01.2022 15:49
In the cartesian system of coordinates $xOy$ consider the variable point $M(x,y)$, ($x,y\in\mathbb{R}$) and the fixed points: $A(-4,-2);B(5,-4);C(2,6);D(5,6)$. $\sqrt{(x+4)^2 + (y+2)^2}=MA;\;\sqrt{(x-5)^2 + (y+4)^2}=MB$; $\sqrt{(x-2)^2 + (y-6)^2}=MC;\;\sqrt{(x-5)^2 + (y-6)^2}=MD$. We observe: $AC=BD=10$. The requested inequality becomes: $MA+MB\le MC+MD+AC+BD\quad(1)$. Applying the triangle inequality in $\triangle{MAC}$ and $\triangle{MBD}$ results: $MC+AC\ge MA$; $MD+BD\ge MB$. Adding, results the inequality $(1)$. $AC\cap BD=\{E\}$, with the coordinates $E(5,10)$. The equality in $(1)$ occurs for $M\equiv E\Longleftrightarrow \begin{cases}x=5;\\y=10.\end{cases}$
26.01.2022 16:37
16.11.2023 13:42
Using Minkowski's inequality, we have $RHS = (\sqrt{(x-2)^2 + (y-6)^2} + \sqrt{6^2 + 8^2}) + (\sqrt{(x-5)^2+(y-6)^2} + \sqrt{0^2 + 10^2}) \ge \sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} = LHS$
16.11.2023 14:18
nchuong1312 wrote: $$RHS = (\sqrt{(x-2)^2 + (y-6)^2} + \sqrt{6^2 + 8^2}) + (\sqrt{(x-5)^2+(y-6)^2} + \sqrt{0^2 + 10^2}) \ge \sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} = LHS$$ Very nice !