For 0 < x < y, show that x < √(xy) < 1/2(x+y) < y?

Answer 1

0 < x < y

x < y Add x to both sides so 2x< x + y ie x < ½(x + y)
Now add y to both sides so x + y < 2y ie ½(x + y) < y

Thus x < ½(x + y) < y ...... (1)

x < y
so x² < xy
so x < √(xy)

also xy < y²
so √(xy) < y

ie x < √(xy) < y ......... (2)

Since 0 < x < y
x - y < 0 (≠0)

So 0 < (x - y)² ( as the square any non-zero real number is positive)
So 0 < x² - 2xy + y²

ie 4xy < x² + 2xy + y² = (x + y)²
so 2√(xy) < (x + y)

ie √(xy) < ½(x + y) ...... (3)

Combining ALL results

x < ½(x + y) < y ...... (1)
x < √(xy) < y ......... (2)
√(xy) < ½(x + y) ...... (3)

x < √(xy) < ½(x + y) < y

Answer 2

I'd start by considering what happens when x and y are equal. In that case, the problem would be expressed as:

x = sqrt(x*x) = 1/2 (2x) = x
and
y= sqrt(y*y) = 1/2(2y) = y

If you make y just a little bit bigger than x, then x^2 < xy, which means the sqrt of (xy) has to be larger than the sqrt (x^2).

If you make y just a little bit bigger than x, then x+y has to be bigger than 2x, which is just x+x. Therefore, 1/2 (x+y) > 1/2 (2x)

If x is just a little bit smaller than y, then 1/2(x+y) has to be less than 1/2 (2y).

You still have to decide if the sqrt(xy) is always smaller than 1/2(x+y). To do that, take the partial derivative of each with respect to y. As y gets bigger than x, how fast does the result change?

The sqrt(xy) can be written (xy)^(1/2). It's derivative is 1/2[sqrt(x)/sqrt(y)] Since x is less than y, then sqrt(x)/sqrt(y) has to be less than one, which means the derivative is less than 1/2.

1/2(x+y) can be written 1/2*x + 1/2*y. The derivative of a constant (which x is in this case) is zero, so you only need to worry about 1/2*y. It's derivative is 1/2.

Since the derivative of sqrt(xy) is less than 1/2, then 1/2(x+y) increases faster than the sqrt(xy) as y increases.

Answer 3

Can't seem to get √(xy) < (1/2)(x+y)...the rest is easy though.

x < √(xy)

Since x < y, √x < √y. knowing that √x = √x, you can multiply both sides by √x getting:

√x * √x < √y * √x

Hence:

x < √(xy)

First part done, next to show that x < (1/2)(x+y):

(1/2)x < (1/2)y

Considering (1/2)x = (1/2)x, add (1/2)x to both sides:

(1/2)x + (1/2)x < (1/2)y + (1/2)x
x < (1/2)(x + y)

Since x < y, x + y < 2y => (x + y)/2 < y

Answer 4

algebra on inequalities works like algebra on equations, except the function applied must be *monotone*. There are two types of monotone functions: nondecreasing functions, where f(x) >= f(y) if x > y; and nonincreasing functions, where f(x) <= f(y) if x > y. The former function applied to both sides of the inequality does not change the truth value of the equation. The latter function may be applied, but the direction of the inequality must be reversed. With this rule and the usual do not divide by zero rule, we may proceed:

0 < x < y

This states that both are non-negative. Multiply throughout by y. Since y non-negative, f(x)=yx is monotone increasing, hense permitted and we do not change direction of the inequality

0 < xy < y^2

take the positive square root all sides. One way to verify that the function is nondecreasing is to look at the graph of f(x)=sqrt(x), and see that it always has positive slope.

0 < sqrt(x*y) < y

Here we have demonstrated one part of the equation. The rest of the answer follows similarly.

Answer 5

take f(u)=u^(a million/2) the thought says that it exist a selection c (0x we prooved a) now for b) only multiply the two members of the inequality a) with the aid of x^(a million/2)