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 $\mathcal{L}=1$.
 A: 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.
A: 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 $AB+BC\geq AC$.
