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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.

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  • $\begingroup$ square it and you can prove this $\endgroup$ – Dr. Sonnhard Graubner Nov 10 '15 at 21:13
  • $\begingroup$ I've gotten to |a-2√a√b+b| ≤ |a−b|, not sure where to go from here? $\endgroup$ – user284408 Nov 10 '15 at 21:16
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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.

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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 $a\ge b$, $a+b-2\sqrt{ab}\ge a-b$.

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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

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

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