Proving the inequality $|\sqrt{a} −\sqrt{b}| \le \sqrt{|a − b|}$

Question

How to prove the following inequality? $$|\sqrt{a} −\sqrt{b}| \le \sqrt{|a − b|}$$ if $a,b >0$

I've gotten to $|a-2\sqrt{a}\sqrt{b}+b | \le |a−b|$, not sure where to go from here.

Answer 1

Assume $0<a<b$ then $$|a−2a \sqrt b\sqrt a +b|≤|a−b|$$ can be rewritten as $$a−2 \sqrt b\sqrt a +b≤a−b \iff 2b<2 \sqrt b\sqrt a,$$ which is true given that $0<a<b.$ Further by symmetry, the case $0<b<a$ also holds. Also, the case when $a=b$ is trivial.

Answer 2

HINT:

Without loss of generality, assume that $a\ge b\ge 0$ so that $|\sqrt{a}-\sqrt{b}|=\sqrt{a}-\sqrt{b}$ and $\sqrt{|a-b|}=\sqrt{a-b}$.

Then note that since $a\ge b$, $a+b-2\sqrt{ab}\ge a-b$.

Answer 3

Squaring both sides =>

LHS = (√a - √b)(√a - √b)

RHS = a - b

Dividing and multiplying LHS by √a + √b =>

LHS = RHS * $\frac{√a - √b}{√a + √b}$

Let $\frac{√a - √b}{√a + √b}$ = k

Its clear that k<=1

Hence, LHS ≤ RHS

Answer 4

Hint: The square root function is concave and vanishes at $0$; therefore $\sqrt{x+y}\le\sqrt{x}+\sqrt{y}$ for $x,y\ge 0$.