Good solution. I'm posit and alternative approach to showing (y - x)/2√y < √y - √x, where 0 < x < y Assume by way of contradition that (y - x)/2√y ≥ √y - √x, where 0 < x < y So (y - x)/(√y - √x) ≥ 2√y, as √y - √x > 0 and 2√y > 0 So (√y + √x) ≥ 2√y, as (√y + √x)(√y - √x) = (y - x) So √y + √x ≥ 2√y = √y + √y So √x ≥ √y So x ≥ y, which contradicts x < y. So, we conclude (y - x)/2√y < √y - √x, where 0 < x < y
Good solution. I'm posit and alternative approach to showing (y - x)/2√y < √y - √x, where 0 < x < y
Assume by way of contradition that
(y - x)/2√y ≥ √y - √x, where 0 < x < y
So (y - x)/(√y - √x) ≥ 2√y,
as √y - √x > 0 and 2√y > 0
So (√y + √x) ≥ 2√y,
as (√y + √x)(√y - √x) = (y - x)
So √y + √x ≥ 2√y = √y + √y
So √x ≥ √y
So x ≥ y, which contradicts x < y.
So, we conclude
(y - x)/2√y < √y - √x,
where 0 < x < y
You're waooooo
How did you choose f(x) to be the square root of x?
Because the middle term in the inequality is sq(y)-sq(x)
Well proved
Can you please say why the interval will be from (0,a)
Because x>0 is given in the question
what happen between 5.43 - 6.03 ie why you change the sign of the inequality
I changed the location of the endpoints of inequality
Why a=x nd b=y
To change the name of variables to be as mentioned in the question