Prove $|x+y| \leq (1+x^2)(1+y^2)$.

Question

Prove $|x+y| \leq (1+x^2)(1+y^2)$.

I don't know if this is true, but I'm trying to build a proof with this property. I know this is true when both $x,y \geq 1$ and $x,y =0$ as well as $\vert x+y\vert < 1$. But the case when $1 < \vert x+y\vert < 2$ is what's giving me trouble. If this property is false, could you provide a counterexample?

(Sorry for the tag, not sure what else to put it under).

Answer 1

By C-S $$\sqrt{(1+x^2)(y^2+1)}\geq\sqrt{(x+y)^2}=|x+y|.$$

Also, by AM-GM and C-S we obtain: $$(x^2+1)(y^2+1)=x^2y^2+x^2+y^2+1\geq x^2+y^2+1\geq$$ $$\geq2\sqrt{x^2+y^2}=\sqrt2\sqrt{(1^2+1^2)(x^2+y^2)}\geq\sqrt2|x+y|\geq|x+y|$$

Answer 2

Denote by $\langle u,v \rangle = \sum_{j=1}^2 u_j v_j$ the scalar product of two vectors $u=(u_1,u_2)$, $v=(v_1,v_2) \in \mathbb{R}^2$. As

$$x+y = \langle \begin{pmatrix}x \\ 1\end{pmatrix}, \begin{pmatrix} 1 \\ y \end{pmatrix} \rangle$$

it follows from the Cauchy Schwarz inequality that

$$|x+y| \leq \sqrt{x^2+1} \sqrt{y^2+1}.$$

Since $\sqrt{z} \leq z$ for any $z \geq 1$, this implies that

$$|x+y| \leq (x^2+1) (y^2+1).$$

Answer 3

Consider that $0 < \frac{1}{2} + (x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 + x^2y^2$ since every term on the RHS is $\geq 0$.