Proving a Quick Trigonometric Inequality

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #ProvingInequalities

EXPLORE 😎:

PLAYLISTS 🎵 :

[mrphlip]

To simplify, set u=sin²x, now our formula is just
1/u + 1/(1-u) = 1/(u-u²)
Now, consider u-u²: this is a quadratic, with roots at 0 and 1, so it is maximised at ½, with maximal value ¼.
This means that 1/(u-u²) is minimised at ½, with minimal value 4. (for u between 0 and 1)

[יונתןסברדלוב-יק]

You can easily substitute sin(x)^2=t and then you need to prove that for t between (0, 1) 1/t+1/(1-t)>=4

[theloganator]

This is even easier with HM-AM inequality!
The arithmetic mean is greater than or equal to the harmonic mean.
(x1+x2)/2 ≥ 2/(1/x1 + 1/x2)
Multiply both sides by 2 and by the denominator on the right,
(x1+x2)(1/x1 + 1/x2) ≥ 4
Set x1 = sin^2x, x2 = cos^2x, and notice that then x1+x2 = 1. The result is
1/sin^2x + 1/cos^2x ≥ 4
as desired.

[paoloputrino]

I don't speak english very well, but I want to give you a more quicky and simple solution, that consider only the fact sin²x + cos²x = 1.
If we sobstitute cos²x with 1 - sin²x, and we develop the expression with algebric step, the expression became
1/(sin²x * (1 - sin²x))
For simplicity we call sin²x = u. We want to demonstrate that
1/(u * (1 - u)) ≥ 4 It is equivalent to
u * (1 - u) ≤ ¼ that is equivalent to
u² - u + ¼ ≥ 0. The expression we got is a square
(u - ½)² ≥ 0 and a square is always positive or equal to zero.
So we proved the initial expression, since it was equivalent to this.

[tommasonobili]

You can also consider a more intuitive solution, csc^2 and sec^2 intersect always at a point with the same y value and you can easily see that that point is the minimum of the function csc^2 x + sec^2 x. This point occurs at pi/4 + k.pi and csc^2 (pi/4)=sec^2 (pi/4)=2 thus sec^2 +csc^2 >=4

[RajeshYadav-nbwp]

AM≥HM

(sin²x+cos²x)/2≥2/(1/sin²x +1/cos²x) implies the relation you needed. Easy but still good one👍

[WahranRai]

Another way to demonstrate inequality is to think about extrema (minimum, maximum) of function.
f(x) = 1/sin(x)^2 + 1/cos(x)^2
f'(x) = 2sin(x)/cos(x)^3 - 2cos(x)/sin(x)^3 then f'(x) = 0 for x=45 degre .
Study of the sign of f'(x) will show that f(x) is minimum for x=45 and the value of this minimum is 1/sin45^2 + 1/cos(45)^2 = (2/sqrt(2))^2 + (2/sqrt(2))^2 = 4

[MrLidless]

By symmetry, sin x = cos x = sqrt(1/2) gives a minimum. Therefore >=4. Done.

[vicebubble]

Cauchy–Bunyakovsky-Schwarz inequality - easiest way

[justariver]

1/sin²x +1/cos²x >=(1+1)²/(sin²x+cos²x) =4. i think this is the easiest way by using cauchy schwarz inequality

[ahmadabdalraheem]

An easier thing
(Cosx)²+(Sinx)²>=(2SinxCosx)²
1>=(Sin(2x))²
We know that
Sin²(something)+Cos²(the same thing)=1
So Sin²(2x)=1- Cos²(2x)
1>=1- Cos²(2x)
0>= - Cos²(2x)
Move to the other side
Cos²(2x)>=0
Which is true, Cos²(2x) is always positive or equal to zero because it's squared.

[جمالالدين-مف]

We can use the formula x + 1/x >= 2 other method

[JohnRandomness]

On the left, replace both ones with cos²x + sin²x. Then you get 1/tan²x + 1 + 1 + tan²x = 2 + u + 1/u. The minimum value of u + 1/u: set its derivative equal to zero: 1 - 1/u² = 0 ==> u = 1 (or -1, not this case). That means the minimum value of u + 1/u is 2. So the minimum value of 1/sin²x + 1/cos²x is 4.

[siddharthabhattacharya]

Am≥Hm directly implies the result

(1/sin²(x)
So required is ≥4

[darkomarkovic]

Maybe we can use inquality (x+y)/2>=(xy)^1/2

[ThePeterDislikeShow]

Does this still work if you allow x to be a complex number?

[prabu]

Thought this problem is simple, it has sparked health and interesting discussion. The ideas and the methods, discussed in the comments section, to solve the same problem in different ways is interesting.

[ahaanroychowdhury]

Multiply both sides by sin squared x cos squared x.
You get sin squared x + cos squared x > 4 sin squared x cos squared x
A.k.a 1> (2sinxcosx)^2
A.k.a 1>( sin2x)^2
Max value of (sin2x)^2 is 1 cuz the max value of sin2x is one, so you can prove it this way I think.

[gdtargetvn]

I have another solution which might be shorter and of course, simpler.

- First, we need to prove that for 2 non-negative number x and y, we always have 1/x+1/y >= 4/(x+y)
So, by using AM-GM, we have:
x+y >= 2√(xy)
(x+y)^2 >= 4xy
(x+y)^2 * (xy)/(xy) >= 4xy(x+y)/(x+y)
Because x, y>0, we can divide both sides by (x+y)/(xy)
(x+y)/(xy) >= 4/(x+y)
1/x+1/y >= 4/(x+y)
The equality is only attained if and only if x = y

- Now, it's obvious that sin²x > 0 and cos²x > 0, so, by using the inequality we have proved above, we have 1/(sin²x)+1/(cos²x) >= 4/(sin²x+cos²x), which is basically the inequality we need to prove at first, since sin²x+cos²x=1.

[lazaremoanang]

Ok, I said earlier that I'll be back. I was thinking about another method before going. When studying the function x---> tan²x+cotan²x-2, you can see that any value is at least equal to 0 so tan²x+cotan²x >=2 <=> 1+tan²x+1+cotan²x >=4 <=> 1+cotan²x+1+tan²x >=4 <=> (1/sin²x)+(1/cos²x) >=4, that's all folks.