# Show that for any $x,y\in\mathbb{R}$ it is $\sqrt{1+x^2} \le \sqrt{1+y^2}+|x-y|$

Problem: "Show that for any $$x,y\in\mathbb{R}$$ it is $$\sqrt{1+x^2} \le \sqrt{1+y^2}+|x-y|$$."

My textbook solves this with the mean value theorem, I tried another solution; can someone check it, please?

My solution: the inequality is equivalent to $$\sqrt{1+x^2} -\sqrt{1+y^2}\le|x-y|$$. Since for any $$u \in \mathbb{R}$$ it is $$u \le |u|$$, it is $$\sqrt{1+x^2}- \sqrt{1+y^2}\le \left|\sqrt{1+x^2}- \sqrt{1+y^2}\right|$$; thus, proving $$\left|\sqrt{1+x^2}- \sqrt{1+y^2}\right| \le |x-y|$$ implies the inequality.

The inequality $$\left|\sqrt{1+x^2}- \sqrt{1+y^2}\right| \le |x-y|$$ is equivalent to showing that the function $$f:\mathbb{R} \to \mathbb{R}$$ defined by $$f(t)=\sqrt{1+t^2}$$ is Lipschitz continuous with Lipschitz constant $$\mathcal{L}=1$$. A differentiable function is Lipschitz continuous if and only if its derivative is bounded, and its Lipschitz constant is the supremum of the modulus of the derivative. Since $$f$$ is differentiable on $$\mathbb{R}$$ with $$f'(t)=\frac{t}{\sqrt{1+t^2}}$$ From $$\sqrt{1+t^2} \ge |t|$$, it follows that $$|f'(t)|\le 1$$ and so $$1$$ is an upper bound for $$f'$$; since $$\lim_{t \to \infty} f'(t)=1$$, it follows that $$\sup_{t \in \mathbb{R}} \{|f'(t)|\}=1$$ and so $$f$$ is Lipschitz continuous with Lipschitz constant $$\mathcal{L}=1$$.

• An alternative, much easier way, is to draw a picture. The value $\sqrt{1+x^2}$ is the length of the vector $(x,1)\in\mathbb{R}^2$. Similarly for $\sqrt{1+y^2}$ and the point $(y,1)$. Once these two vectors are drawn you're left with $|x-y|$ which just so happens to be equal to the distance between the two points. Draw the points and discover the triangle inequality. Commented Apr 9, 2022 at 13:14

It's correct, but it's essentially the same approach as the mean value theorem: you are using a bound on the derivative to show the inequality. And proving the inequality from the bound on the derivative for a Lipschitz function would involve the mean value theorem.

Here is an algebraic way.

Assume $$x>y$$. Then, as $$t<\sqrt{t^2+1}$$ for all real $$t$$, we have

$$\frac{x+y}{\sqrt{x^2+1}+\sqrt{y^2+1}}\le1$$

Since $$x>y$$, multiplying by $$x-y$$ doesn't change the inequality:

$$\frac{x^2-y^2}{\sqrt{x^2+1}+\sqrt{y^2+1}}\le x-y$$ $$\frac{(x^2+1)-(y^2+1)}{\sqrt{x^2+1}+\sqrt{y^2+1}}\le x-y$$ $$\frac{(\sqrt{x^2+1}+\sqrt{y^2+1})(\sqrt{x^2+1}-\sqrt{y^2+1})}{\sqrt{x^2+1}+\sqrt{y^2+1}}\le x-y$$

$$\sqrt{x^2+1}-\sqrt{y^2+1}\le x-y$$

And $$x-y>0$$, hence

$$\sqrt{x^2+1}-\sqrt{y^2+1}\le |x-y|$$

And if $$x\le y$$, the left hand side is nonpositive, so the inequality is trivially true.

Another approach: consider points $$A(x,0)$$, $$B(y,0)$$ and $$C(0,1)$$ on $$\mathbb{R}^2$$. Then, your inequality is equivalent to the triangle inequality $$AB+BC\geq AC$$.

• Note that this answer was essentially contained in @Kolja's comment (my bad, didn't notice it). Commented Apr 9, 2022 at 14:15