Inequalities Question (1 Viewer)

[theprofitable95]

Consider the function f(x) = 3.sinx / (2 + cosx). Show that f(x) < x for x>0. This was Q16 from a 2012 trial paper, and the solutions used calculus (proof that derivative of x - f(x) is always positive). I was wondering if there was a non-calculus way to prove this. Something I came up with was:

sinx < x
3sinx < 3x

Since 3>2+ cosx>1 for all x>0, we can divide both sides.

3sinx/(2+cosx) < 3x/(2+cosx)
f(x) < 3x/(2+cosx) .... and im not sure how to continue from here. The thing is that 1< 3/(2+cosx) <3, so the only proof I can get from this is that f(x)<3x

 

[Sy123]

 

[braintic]

I think there's an error there.

 

[Sy123]

I think there's an error there.

Yes my bad

 

[seanieg89]

I would be pretty surprised if there was a non-calculus proof of this that wasn't just "calculus in disguise", because you need to look at higher order behaviour of the trig functions (hence the differentiations).

Inequalities like sin(x) < x lose too much information.

 

[seanieg89]

And I count using Taylor series as "calculus in disguise" here, because the trig functions are not defined by these series in MX2.

 

[Sy123]

How about this:

 

[braintic]

As it turns out, that is correct.
But given that x is increasing and arctan(3/x) is decreasing, how did you know that without calculus?

(But it took me a while to decipher what you had written, so perhaps I have missed something)

 

[seanieg89]

How about this:

Why does it suffice to show that cos f(x) doesn't attain a minimum on that small interval? Can't your entire LHS conceivably be negative without this occurring?

 

[Sy123]

Yea I can't see how to resolve that without doing some calculus in disguise

 

[theprofitable95]

Ok so for more 'complicated' single variable inequalities we should generally use calculus. Thanks for the help.

 

[braintic]

Ok so for more 'complicated' single variable inequalities we should generally use calculus. Thanks for the help.

When you have function that 'don't go together'.

Like trig and polynomials in this question.

 

[eal]

Maths questions
1. Sketch f(x) and prove the following inequalities:
a. F(x) = 1/x : 1 + ½ + 1/3 + … + 1/(n-1) (greater than or equal to) ln(n)
b. F(x) = 1/(x)^1/2 : 1/(2)^1/2 + 1/(3)^1/2 + … + 1/(n)^1/2 (less than or equal to) 2((n)^1/2 -1)
2. Prove that 2(a^3 + b^3 + c^3) (greater than or equal to) a^2b + ab^2 + a^2c + ac^2 + b^2c + c^2b for positive real numbers
3. Prove that a^2 + b^4 + c^6 (greater than or equal to) ab^2 + b^2c^3 + c^3d^4 + d^4a for all real numbers HINT: ab^2 = (a^2b^4)^1/2

4. Prove that xyz (less than or equal to) 1 supposing that (1+x)(1+y)(1+z) = 8 and using a^3 + b^3 + c^3 (greater than or equal to) 3abc and x + y + z (greater than or equal to) 3((xyz)^1/3

I really need some help with the solutions to these! Thanks!

 

[mathing]

Maths questions
1. Sketch f(x) and prove the following inequalities:
a. F(x) = 1/x : 1 + ½ + 1/3 + … + 1/(n-1) (greater than or equal to) ln(n)
b. F(x) = 1/(x)^1/2 : 1/(2)^1/2 + 1/(3)^1/2 + … + 1/(n)^1/2 (less than or equal to) 2((n)^1/2 -1)
2. Prove that 2(a^3 + b^3 + c^3) (greater than or equal to) a^2b + ab^2 + a^2c + ac^2 + b^2c + c^2b for positive real numbers
3. Prove that a^2 + b^4 + c^6 (greater than or equal to) ab^2 + b^2c^3 + c^3d^4 + d^4a for all real numbers HINT: ab^2 = (a^2b^4)^1/2

4. Prove that xyz (less than or equal to) 1 supposing that (1+x)(1+y)(1+z) = 8 and using a^3 + b^3 + c^3 (greater than or equal to) 3abc and x + y + z (greater than or equal to) 3((xyz)^1/3

I really need some help with the solutions to these! Thanks!

1a) Look at the graph of it and the unit rectangles formed, then take the integrals
1b) As above

2) Prove then use a^3 + b^3 >= a^2 + ab^2

3) Looks like a typo, no d on LHS

4) apply (x + y + z) >= 3(xyz)^(1/3) with x -> 1 + x etc. we get (3 + x + y + z) >= 6 so we have x + y + z >= 3
Now apply (x + y + z) >= 3(xyz)^(1/3) and we know (xyz)^(1/3) <= 1 so xyz <= 1, this assumes xyz > 0 but it has to be otherwise that (1+x)(1+y)(1+z) = 8 couldn't hold.