# Subadditivity of square root function [duplicate]

Any hint to prove

$$\sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0$$

## marked as duplicate by Martin Sleziak, Claude Leibovici, user91500, JonMark Perry, Joel Reyes NocheAug 7 '16 at 6:18

• Simply square both sides. – Aditya Anand Dec 25 '15 at 17:12

Notice that $$\sqrt{x+y}\le \sqrt{x}+\sqrt{y}\Leftrightarrow x+y\le x+y+2\sqrt{xy}.$$

Which clearly holds as $x,y\ge{0}$, and the function $x\rightarrow x^2$ is monotonic and equality occurs when one of $x$ or $y$ is $0$.

• So, assuming the result that you want to prove, you are able to show something that is clearly true. (I hope you see the problem.) – Andreas Rejbrand Dec 26 '15 at 0:19
• This would be correct if you replaced $\implies$ with $\iff$ (and, perhaps, pointed out that the equivalence is true because $x \mapsto x^2$ is strictly monotone increasing, and thus order-preserving, for non-negative real $x$). – Ilmari Karonen Dec 26 '15 at 1:08

Another:

$x,y \geq 0$ then the product $\sqrt{x} \sqrt{y}$ have sence and now $$\sqrt{x + y} \leq \sqrt{x + 2\sqrt{x}\sqrt{y} + y} = \sqrt{(\sqrt{x} + \sqrt{y})^2} = \sqrt{x} + \sqrt{y}$$

I'll try the hard way and use calculus.

Let $f(y) = \sqrt{x+y} - \sqrt{x} - \sqrt{y}$.

$f(0) =0$.

$f'(y) =\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{y}}$. Since $x+y \ge y$, $\sqrt{x+y} \ge \sqrt{y}$ so $\frac1{2\sqrt{x+y}} \le \frac1{2\sqrt{y}}$.

Therefore $f'(y) \le 0$ for all $y$.

Since $f(0) = 0$, $f(y) \le 0$ for all $y$.

Note that this can be used to show that if $g(y)$ is increasing and $g'(y)$ is decreasing then $g(x+y) \le g(x)+g(y)$.

Similarly, if $g'(y)$ is increasing, $g(x+y) \ge g(x)+g(y)$.

The inequality is homogeneous, hence we may assume $y=1$ without loss of generality.

So we just have to prove that $\sqrt{x+1}-\sqrt{x}\leq 1$, pretty easy: $$\sqrt{x+1}-\sqrt{x} = \frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \frac{1}{\sqrt{x+1}+\sqrt{x}}\leq 1.$$

• May I ask what it means for an inequality to be homogeneous? – Kari Dec 25 '15 at 17:39
• @Khallil: that if we replace $x$ and $y$ respectively by $\lambda x$ and $\lambda y$ (with $\lambda > 0$) the inequality is unchanged. – Jack D'Aurizio Dec 25 '15 at 17:47