Problem with $A.M.geq G.M.$ Inequality

Use $T_2$'s lemma to get, $$1=x^2+y^2geq frac{(x+y)^2}{2}$$

Using the Cauchy-Schwarz inequality we have $$(x+y)^2 leqslant (1+1)(x^2+y^2) = 2(x^2+y^2) = 2.$$ So $$|x+y| leqslant sqrt{2},$$ or $$ -sqrt 2 leqslant x + y leqslant sqrt 2.$$

$$x^2+y^2=1\(x+y) ^2=1+2xytag{1}$$ Now, you've got that $$frac{1}{2}ge xy\ frac{1}{2}ge frac{(x+y) ^2-1}{2}tag{from (1)}\ (x+y) ^2le 2\ -sqrt 2le x+yle sqrt 2$$

Another way: set $x=costheta, y = sintheta,$ then $$ x+y=costheta+sintheta=sqrt{2}cos(theta-pi/4) $$

Proceeding from your approach, you had $$2xy leq 1$$ Adding $x^2 + y^2$ to both sides, $$implies x^2+2xy+y^2 leq 1 + x^2 + y^2$$ $$implies (x+y)^2 leq 1 + 1$$ And you're done.

Using Brahmagupta-Fibonacci Identity or directly

$$(x+y)^2+(x-y)^2=(1^2+1^2)(x^2+y^2)$$

$$2(x^2+y^2)-(x+y)^2=cdotsge0$$