New Year Countdown - 7

$\large\color{#69047E}{Please~ upvote~ this~ solution ~if ~you ~like ~it}$

First of all by Rearrangement inequality we can assume $\large a>b>c$

$\large \Rightarrow \therefore a^2+b^2+c^2\geq ab+bc+ca$

$\large \Rightarrow 1 \geq ab+bc+ca$

Means $\large Y=1$

And we know that $\large (a+b+c)^2 \geq 0$ $~~$ (As $a,b,c$ are reals)

$\large \therefore$ $\large a^2+b^2+c^2+2ab+2bc+2ca \geq 0$

$\large \therefore$ $\large 1+2(ab+bc+ca) \geq 0$

$\large \therefore$ $\large (ab+bc+ca) \geq \dfrac{-1}{2}$

$\large \therefore X=\dfrac{1}{2}=0.5$

$\Rightarrow \huge \color{#3D99F6}{X+Y=\boxed {1.5}}$

nous avons: ab+bc+ca=((a+b+c)^2-a^2-b^2-c^2)/2 donc ((a+b+c)^2-1)/2=ab+bc+ca par caushy shwarz:3≥(a+b+c)^2≥0 finalement:1≥ab+bc+ca≥-1/2 alors:x+y=1.5

$\text{https://brilliant.org/problems/question-11-2/}$ is a similar problem.