# Trig Inequality (partial solutions included)

• lovemake1
That's the whole point of the exercise.In summary, by given 0 <= a <= b, we can show that sqrt(ab) <= (a + b)/2 is true by first proving that 0 <= a^2 - 2ab + b^2 is also true for all nonnegative values of a and b. This is because the square root function only outputs positive real numbers, so a*b must be greater than 0. Thus, we can conclude that sqrt(ab) is always less than or equal to (a + b)/2 for all 0 <= a <= b.
lovemake1

## Homework Statement

Given 0 <= a <= b

show that,

a <= sqrt(ab) <= (a+ b / 2) <= b

## Homework Equations

a * b <= a^2 / a*b <= a* a

## The Attempt at a Solution

I think i know where I am going but i wanted to make sure if its correct so far.

So we know that the order of least to greatest, 0 -> a -> b

and the first part of inequality states that a <= (Sqrt)ab
so i take the sqrt and squre the left side so it makes a ^2.

The inequality is now a^2 <= ab
and this is true because a < b. and a^2 = a * a
there fore a* a is smaller than a * b.

The second part is [sqrt(ab) <= a+ b/ 2]
Sqrt(ab) smaller or equal to (a + b) / 2.

Solution: square both sides, ab <= [ (a+b)/ 2 ]^2
and this gives 4ab <= (a+b)^2
4ab <= a^2 + 2ab + b^2

and last part

a + b / 2 <= b
multiply by 2 to both sides.
a+b <= 2b

and sinec a < b and
a + b <= b * b

is this correct? am i allowed to do what i just did ? please help me

Last edited:
What does this mean -- 0 <\ a <\ b?

Why are the slashes there?

For less than, use <
For less than or equal, use <=
Similar for greater than and greater than or equal.

Sorry about that, fixed to <=

Can you show that (a + b)/2 is between a and b?
Can you show that sqrt(ab) <= (a + b)/2?

a) (a+b)/2 between a and b

yes, setting up an equality so that a <= (a+b)/2 <= b
which is same as 2a <= a+b <= 2b

this prooves that (a+b)/2 is between a and b2) sqrt(ab) <= (a+b)/2

square both sides to remove sqrt from ab.

(ab) <= [(a+b)/2]^2
(ab) <= [ (a*b)^2/4]

4ab <= (a*b)^2

is it ok to leave it as 4ab <= (a*b)^2
or should i be expanding so its 4ab <= a^2 + 2ab + b^2 or even go beyond this??

lovemake1 said:
a) (a+b)/2 between a and b

yes, setting up an equality so that a <= (a+b)/2 <= b
which is same as 2a <= a+b <= 2b

this prooves that (a+b)/2 is between a and b
I wouldn't do it that way.
You're given that 0 <= a <= b.
a = (a + a)/2 <= (a + b)/2, since a/2 <= b/2 (which is because a <= b).

It's about as easy to show that (a + b)/2 < b.
lovemake1 said:
2) sqrt(ab) <= (a+b)/2

square both sides to remove sqrt from ab.

(ab) <= [(a+b)/2]^2
(ab) <= [ (a*b)^2/4]
The step above is not obvious (to me). I would expand the (a + b)/2 part.
lovemake1 said:
4ab <= (a*b)^2

is it ok to leave it as 4ab <= (a*b)^2
or should i be expanding so its 4ab <= a^2 + 2ab + b^2 or even go beyond this??

im really confused, did you just want me to make it 4ab <= a^2 + 2ab + b^2
and subtract 2ab from both sides to make 2ab <= a^2 + b^2 ?

No, subtract 4ab from both sides.

oh, just make " 0 " on the otherside? (Didnt know this was allowed)
this indefinetly proves that it is greater than left side right?

so i would end up with 0 <= a^2 - 2ab + b^2
making the right side greater.

Sure, you can always add or subtract the same amount from both sides of an equation or inequality.

For 0 <= a^2 - 2ab + b^2, for what values of a and b? That's crucial.

do i have to find what values a and b have to be ?
a^2 - 2ab + b^2 = (a-b)^2
or just leave it at 0 <= a^2 - 2ab + b^2 to prove sqrt(ab) <= (a+b)/2

from the question it seems like their only asking for the 0 <= a^2 - 2ab + b^2 but also i think i should prove further to see if they are indeed greater than 0.

(a-b)^2 how would this be interpreted as number or equality?
confused.

What you have is a sequence of equivalent inequalities, starting with
sqrt(ab) <= (a + b)/2
and going to 0 <= a^2 - 2ab + b^2

If you know which values the last inequality is true for, you know the values the first one is true for.

how do i find the values?
im slowly getting lost.

i have to prove that
a^2 - 2ab + b^2 is greater or equal to 0.
but how would i do that?

(a-b)^2 >= 0
any number of A that is smaller than B would give a positive real number.

sry I am lost here

For this problem a - b can be negative or zero, but when you square a - b, you get a number that is always at least zero.

So (a - b)^2 >= 0 for all real numbers a and b. This means that (a + b)/2 >= sqrt(ab) for all nonnegative real numbers a and b. The square root isn't a real number if a and b are opposite in sign, but you're given that both a and b are >= 0.

ah.. i see the problem.
a * b cannot be a negative number because of the square root.

not sure how it would work out..
ab would always have to > 0

Not a problem. You're given that 0 <= a <= b, so neither one of them can be negative.

oh right, so it does work out afterall.
0 <= a^2 - 2ab + b^2

this would be the correct final stage then?
given 0 < a < b

No, you don't want to end up with 0 <= a^2 - 2ab + b^2. You want to show from this being true for all values of a and b such that 0 <= a <= b, that sqrt(ab) <= (a + b)/2. That's what you want to conclude in this part of the problem.

Part of doing that is to explain why the inequalities sqrt(ab) <= (a + b)/2 and 0 <= a^2 - 2ab + b^2 have the same solution set, at least for 0 <= a <= b.

## What is a trig inequality?

A trig inequality is a type of inequality that involves trigonometric functions, such as sine, cosine, and tangent. It is an equation that compares the values of two trigonometric expressions.

## How do you solve a trig inequality?

To solve a trig inequality, you can use algebraic techniques such as factoring, completing the square, or using the quadratic formula. You can also use trigonometric identities and properties to simplify the inequality and find its solutions.

## What are the common mistakes when solving trig inequalities?

Some common mistakes when solving trig inequalities include forgetting to check for extraneous solutions, making calculation errors, and not considering the restrictions on the trigonometric functions. It is important to carefully check each step of the solution and double-check the final answer.

## Why are trig inequalities relevant in mathematics?

Trig inequalities have many applications in fields such as physics, engineering, and astronomy. They are also important in calculus, as they are used to find the maximum and minimum values of trigonometric functions.

## Can you give an example of solving a trig inequality?

Yes, for example, let's solve the inequality sin(x) > 0. To do this, we first need to find the values of x that make sin(x) greater than 0. Based on the unit circle, we know that sin(x) is positive in the intervals [0, π/2] and [3π/2, 2π]. Therefore, the solution to the inequality is x ∈ [0, π/2] ∪ [3π/2, 2π].

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