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

Question

Question:

If $x^2+y^2=1$, prove that $-{\sqrt2}\leq x+y \leq\sqrt2$

My approach: $$\frac{x^2+y^2}{2}\geq\sqrt{x^2y^2}$$ $$ \frac12\geq xy$$ $$\frac{-1}{\sqrt2}\leq\sqrt{xy}\leq \frac{1}{\sqrt2} $$ Now how do I proceed from here?

Answer 1

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.

Answer 2

Using Brahmagupta-Fibonacci Identity or directly

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

$$2(x^2+y^2)-(x+y)^2=\cdots\ge0$$

Answer 3

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

Answer 4

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

Answer 5

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 $$ -\sqrt 2 \leqslant x + y \leqslant \sqrt 2.$$

Answer 6

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