New Year Countdown - 8

We have $a^{2}+b^{2}=2$

Let $a+b=S$ $\therefore ab=\frac{S^{2}-2}{2}$

So, By Vieta's relations, $a,b$ are the roots of the monic polynomial:

$\large{P(x)=x^{2}-Sx+\frac{S^{2}-2}{2}}$

since $a,b\in\mathbb{R}$ , the discriminant $\Delta≥0$

checking $\Delta=S^{2}-2.(S^{2}-1)≥0$

$\implies \large{S\in [-2,+2]}$

our answer $=\large{\boxed{4}}$

By cauchy we have

${(1\times a)+(1\times b)}^2\leq {(a^2+b^2)(2)}$

$\therefore maxima (a+b)=2$

similarly we can get

$minima~a+b=-2$ at $(a,b)=(-1,-1)$

$\therefore X+Y \Rightarrow \huge\boxed{4}$

Without the loss of generality, we can assume $x= \sqrt{2}sin\quad\theta$ and $y= \sqrt{2}cos \quad \theta$ .

Then, $(a+b)= \sqrt{2}(sin \quad \theta +cos\quad \theta)$ .

Now the maximum value of $(sin\quad \theta +cos \quad \theta )$ is $\sqrt{2}$ and its minimum value is $-\sqrt{2}$ .

Thus $(a+b)\epsilon [-2,2]$ i.e., X=2 and Y=2.

So, $\boxed{X+Y=4}$ .

using AM>=GM ; we have (a^2+b^2)/2>=ab ==> ab<=1 i.e ab=(a+b)^2 /2 -1 <=1 ==> (a+b)^2<=4 ==> -2<=a+b<=2 thus x+y as stated in question comes out to be 2+2=4

PARAMETRIC WAY a=sqrt2 cost , b=sqrt2 sint, hence a+b=sqrt2 cost+sqrt2 sint = 2sin(45+t) which implies max/min = 2/-2 bcoz range of sint is [-1,1] . hence x+y=4

By quadratic mean arithmetic mean inequality, we find that a+b<=2 if a,b are positive. This also means that if both a and b are negative, a+b>= -2 If one of them is negative, and the other is positive, say the numbers are x and -y, then x+(-y)<= x+y<=2. On the other hand, x+(-y)>=(-x) + (-y) =-(x+y)>=-2 .

So X+Y=4