Prove that $\sqrt{c-d} \ge \sqrt{c} - \sqrt{d}$ for $c \geq d \geq 0$

Question

I'm new to math and struggle with a probably pretty easy problem. How do you prove the inequality in the titel? Here's what I got: $$\sqrt{c-d} \geq \sqrt{c - 2 \sqrt{c} \sqrt{d} + d}=\sqrt{ ( \sqrt{c} - \sqrt{d} ) ^2 } = \sqrt c - \sqrt d$$

But I fail to understand the jump from $\sqrt{c - 2 \sqrt{c} \sqrt{d} + d} \ $ to $\sqrt{c-d}$

Thanks for your help :)

Answer 1

You might easier show

• $\sqrt{c-d} + \sqrt{d} \ge \sqrt{c}$

This is quickly done by squaring:

$$\left(\sqrt{c-d} + \sqrt{d}\right)^2 = c-d + d +2\sqrt{c-d}\sqrt{d} \geq c$$

Answer 2

Hint: Squaring the given inequality we get $$-2d\geq -2\sqrt{cd}$$

Answer 3

$c-d=(√c-√d)(√c+√d) \ge (√c-√d)(√c-√d);$

$c-d \ge (√c-√d)^2;$

$\sqrt{c-d} \ge √c-√d.$

Note: $√c+√d > √c-√d >0.$

hence

$(√c-√d)(√c+√d) >$

$(√c-√d)(√c-√d)$.