Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$

Question

Prove that if $0 < a < b$ then $$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$

This is what I have so far:

square both sides to get $a + b -2\sqrt{ab} < b-a$

subtract $b$ from both sides $a-2\sqrt{ab} < -a$

add $a$ to both sides $2a-2\sqrt{ab} < 0$

than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$

divide by $2$ and get $a < \sqrt{ab}$

Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$.

Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$

I'm confused as to where I went wrong and how to fix this.

Answer 1

You just need to note that every step you did is reversible. For the conclusion, from $a\le\sqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.

More simply, set $x=\sqrt{a}$ and $y=\sqrt{b}$. You need to show that $$ y-x<\sqrt{y^2-x^2} $$ Since $y>x$, this is equivalent to $$ (y-x)^2<y^2-x^2 $$ that is, $$ y^2-2xy+x^2<y^2-x^2 $$ which in turn is equivalent, with simple algebraic simplifications, to $$ x^2<xy $$ which is true because $x<y$ and so $x^2<xy$.

Answer 2

It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.

Assuming $0 < a < b$ you have $2\sqrt{a(b-a)}>0$, hence $$2\sqrt{a(b-a)}+a+(b-a) > b.$$ The left hand side is $(\sqrt{b-a}+\sqrt{a})^2$, so $$(\sqrt{b-a}+\sqrt{a})^2 > \sqrt{b}^2$$ and since both sides are positive you can take the square root getting $$\sqrt{b-a}+\sqrt{a} > \sqrt{b},$$ which is your claim.

Answer 3

It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:

We recall that

$0 < a < b; \tag 0$

then, note that

$b = a + (b - a) < a + 2\sqrt a \sqrt{b - a} + (b - a) = (\sqrt a + \sqrt{b - a})^2; \tag 1$

thus,

$\sqrt b < \sqrt a + \sqrt{b - a}, \tag 2$

whence

$\sqrt b - \sqrt a < \sqrt{b - a}. \tag 3$

Answer 4

The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.

(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])

Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.

$0 < a < b$ means that $a\cdot a < a\cdot b < b\cdot b$ and so $a^2 < ab < b^2 $ and $a = \sqrt a^2 < \sqrt {ab} < \sqrt b^2 = b$.

So $a < \sqrt {ab}$

$2a < 2\sqrt {ab}$

$ - 2\sqrt{ab} +2a< 0$

$ -2\sqrt{ab} +a< -a$

$b - 2\sqrt{ab} + a < b - a$

$(\sqrt b -\sqrt a)^2 < b-a$

$|\sqrt b - \sqrt a| < \sqrt{b-a}$.

But as $b > a > 0$ then $\sqrt b > \sqrt a$ and $|\sqrt b - \sqrt a| = \sqrt b - \sqrt a$.

So

$\sqrt b - \sqrt a < \sqrt{b-a}$.