proving an inequality is always interesting because there are 3 main strategies I adopt. I highlighted them in the video and used two of them in combination. Get merch here: rb.gy/cya1qk
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I am Lithuanian and had this exact equation on a specific test in school. Our teacher found it interesting so added it, but issue was this was 9th grade and we just started learning these algebra equations. Makes more sense now when explained.
I think at 10:00, it is easier to factor ab out, cancel out a^2+b^2, and also cancel 2 on right. Result is (a^2+b^2)/2 >= ab, which is true as shown earlier.
Nice problem and video! A couple of minor issues with the proof given: 1) At 5:47 it is concluded that for positive a,b: (a^2+b^2)/2 >= ab In fact this is true for all reals a,b and indeed this is assumed and needed later at 10:27. To prove this inequality for all reals a,b: (a-b)^2 ≥ 0 a^2 - 2ab + b^2 ≥ 0 (a^2 + b^2)/2 ≥ ab 2) At 6:16 you square the above inequality. This step is guaranteed to be true only if the LHS and RHS are positive. However ab is not always positive and in general it is not valid to say that p>q implies p^2>q^2. For example: 1>-10 is true but 1^2>(-10)^2 is false. As it happens, in this case we are okay because the LHS and RHS are not independent values: [(a^2 + b^2)/2]^2 ≥ (ab)^2 (a^4 + 2(ab)^2 + b^4) ≥ 4(ab)^2 a^4 - 2(ab)^2 + b^4 ≥ 0 (a^2 - b^2)^2 ≥ 0 This last line is clearly true meaning that the squared inequality holds. An alternative proof of the original inequality: 2(a^4 + (ab)^2 + b^4) ≥ 3(ba^3 + ab^3) 2[(a^2 + b^2)^2 - (ab)^2] ≥ 3ab(a^2 + b^2) 2[((a+b)^2 - 2ab)^2 - (ab)^2] ≥ 3ab((a+b)^2 - 2ab) 2((a+b)^2 - 2ab)^2 - 2(ab)^2 ≥ 3ab(a+b)^2 - 6(ab)^2 2((a+b)^4 - 4ab(a+b)^2 + 4(ab)^2) - 2(ab)^2 ≥ 3ab(a+b)^2 - 6(ab)^2 2(a+b)^4 - 11ab(a+b)^2 + 12(ab)^2 ≥ 0 (2(a+b)^2 - 3ab).((a+b)^2 - 4ab) ≥ 0 [*] (a+b)^2 - 4ab = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a-b)^2 ≥ 0 2(a+b)^2 - 3ab = 2(a^2 + 2ab + b^2) - 3ab = 2(a^2 + b^2) + ab Let a=r.cos(t) and b=r.sin(t): = 2r^2(cos^2(t) + sin^2(t)) + r^2.sin(t).cos(t) = r^2(2 + sin(2t)/2) ≥ r^2(2 - 1/2) = (3/2)r^2 ≥ 0 So the 2 terms on the LHS of [*] are indeed non-negative.
Thanks for solving and explaining this math problem. I was really stuck on it. Turns out that you're not supposed to replace a^2+b^2 with m and ab with n as I did.
Thank you for the nice problem and its solution. I solved it in a different way, thus maybe worth sharing: If one of a and b is equal to 0, then the RHS of the inequality is 0 and the inequality is true, so we can from now on assume that a*b 0. If one of a and b is negative, than the RHS is either negative and the inequality is true or the other value is also negative and the RHS is positive and this is equivalent to the case where both a and b are positive, and we cannot say at this point whether the inequality is true, so it suffices to show that the inequality is true for a and b both positive. Under this assumption one can divide both sides of the inequality by a^2*b^2 and obtain the equivalent inequality ((a/b)^2+1+(b/a)^2)/3 = ((a/b+b/a)^2-1)/3 >= (a/b+b/a)/2, which after moving the RHS to the LHS is equivalent to ((a/b+b/a-3/4)^2 - 25/16)/3 >= 0. At this place it is very useful to remember that due to AM-GM one has that a/b+b/a >= 2, which leads to ((a/b+b/a-3/4)^2 - 25/16)/3 >= ((2-3/4)^2 - 25/16)/3 = ((5/4)^2 - 25/16)/3 = 0, the inequality following from the fact that the square function is increasing on [0, \infty), which completes the proof.
I like your solution a lot more, but think my solution is correct and solves in a different way (and doesn't assume knowledge of AM-GM inequality) by looking at different domains of a, b. Note that left hand side is always greater than or equal to zero due to even exponents. Right hand side is greater than or equal to zero if a,b are both positive, or both negative otherwise the solution is less than or equal to zero and inequality is trivially true. So we only need to focus on the two cases 1) a >= 0 and b >= 0 without loss of generality assume a >= b and rewrite at a = b c where c>= 1 then substitute and solve in terms of b. After collecting terms you get the inequality b^4 (c^4 + c^2 + 1)/3 >= b^4 (c^3 +c)/2 clearly this inequality holds for all c>= 1 due to quartic power 2) a < 0 and b < 0 without loss of generality assume a = 1 then substitute and solve in terms of b. After collecting terms you get the same inequality as 1
We can manipulate the given inequality into a sort of sum of squares form. First, notice that we only need to prove the inequality for positive a and b, since when a and/or b are negative, the LHS remains the same but the RHS can only decrease or remain the same. Now, for simplicity, we can get rid of the denominators by multiplying the inequality by 6 and move all terms to LHS yielding: 2a^4 + 2b^4 + 2a^2 b^2 - 3a^3 b - 3a b^3 >= 0 This, in turn, can be represented as a (sort of) sum of squares and fourth powers as: (a - b)^4 + (a^2 - b^2)^2 + ab (a-b)^2 >= 0 The "sort of" is due to the fact that there is ab in front of the last square, but since we already justified focusing on non-negative values of a and b in the proof, we are good to go.
The proof seems almost immediate to me: first we increase 2ab with a^2+b^2, obtaining on the second member (a^2+b^2)^2/4 Then the given inequality with the second member increased is easily verified (a perfect square greater than or equal to zero is obtained, clearly always true).
I tried this myself. I tried substituting t = a²+b² and u = ab and solving resulting second order equations for t and u, but it led nowhere. I thought the trick is to show a and b must be real (not complex), so I hoped for the discriminates to help but no such luck. Seeing your solution illustrates how off I was in my attempt.
Tried a similar method. But instead of finding discriminant, I factored out a polynomial 2t^2-2u^2-3tu to get (2t + u)(t-2u)>=0. Knowing that t>=0 and t>=u (from a^2+b^2 = 2ab) it still led me nowhere.
@@Lightseeker1-j5pif you divide by a²b² and note x=a/b you'll obtain a nice inequality where X=x+1/x appears. At the end you have to solve 2X²-3X-2>=0.
I put everything on the same side, and it looks like (a-b)^4, then I keep simplifying the inequation until I got to (a-b)²(2a²+ab+2b²) >= 0. From here we get to a²+b² >-= -ab/2, which I think is always true, but I don't know to explain why.
For the first time in many videos I could not understand your demonstration since I could not figure how the first equality of the proof came from. I guess I have to study more algebra. Thanks anyway.
The arithmetic-geometric mean inequality you cited as justification for some of your steps is true if a and b are positive. How is your proof valid for all real a and b?
This is because a and b are real numbers and the square of any real numbers are positive always. That's why the property AM>=GM holds for a^2 and b^2 as both are greater than or equal to zero. Hope this helps.
I don't agree. You use the same letters a and b for different values. Our initial a and b are not necessary positive numbers. a2 and b2 are posirive, but a and b not necessary. So, I don't see that you don't need to use the absolute value. Anywere, I think it don't affect the result because if one of then is negative, the inequality is more obvious because a3 or b3 will be negative numbers in the right expression and it will be minor.
I mean you use the same letters to enunciate that the arithmetic mean is greather than geometric mean. You should not use a and b for this enunciate, because here a and b are positive numbers, but out original a and b are not. I think you should enunciate the theoreme with another letters. The a and b letters of theoreme are positive numbers, but our initial a and b are not. It's true that a2 and b2 are positive and you can use the theoreme, but when you cancel the square root I think you may use absolute value. I like your videos and I watch all or them. My comments are only with constructive intentions. Sorry for my english.
Two questions: 1. You say that a, b > 0 when employing the AM-GM inequality, yet according to the question, a and b can be negative. 2. At 5:37, you squared the RHS but not the LHS.
@@PrimeNewtons If you mean u = a^2 and v = b^2, it renders all of your subsequent "proof" meaningless. And why are you assuming positive values for a, b, when they can be negative? The AM-GM inequality doesn't hold for negative values.
Bro i am from india i have a challenge bro please solve this if (a+1)(b+1)(c+1)(d+1)=1 and (a+2)(b+2)(c+2)(d+2) =2 and (a+3)(b+3)(c+3)(d+3) =3 and (a+4)(b+4)(c+4)(d+4)= 4 then find (a+5)(b+5)(c+5)(d+5) Btw a i am also mathematics lover
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You can get any Prime Newtons Merch here (35% off in the next 3 days).
Choose your style, size and color. Tanks tops, hoodies, kids, mugs, Baseball hats etc.
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I am Lithuanian and had this exact equation on a specific test in school. Our teacher found it interesting so added it, but issue was this was 9th grade and we just started learning these algebra equations.
Makes more sense now when explained.
I am Lithuanian. I don't remember when I clicked so fast on your video :D
I think at 10:00, it is easier to factor ab out, cancel out a^2+b^2, and also cancel 2 on right. Result is (a^2+b^2)/2 >= ab, which is true as shown earlier.
Nice problem and video! A couple of minor issues with the proof given:
1) At 5:47 it is concluded that for positive a,b:
(a^2+b^2)/2 >= ab
In fact this is true for all reals a,b and indeed this is assumed and needed later at 10:27. To prove this inequality for all reals a,b:
(a-b)^2 ≥ 0
a^2 - 2ab + b^2 ≥ 0
(a^2 + b^2)/2 ≥ ab
2) At 6:16 you square the above inequality. This step is guaranteed to be true only if the LHS and RHS are positive. However ab is not always positive and in general it is not valid to say that p>q implies p^2>q^2. For example: 1>-10 is true but 1^2>(-10)^2 is false. As it happens, in this case we are okay because the LHS and RHS are not independent values:
[(a^2 + b^2)/2]^2 ≥ (ab)^2
(a^4 + 2(ab)^2 + b^4) ≥ 4(ab)^2
a^4 - 2(ab)^2 + b^4 ≥ 0
(a^2 - b^2)^2 ≥ 0
This last line is clearly true meaning that the squared inequality holds.
An alternative proof of the original inequality:
2(a^4 + (ab)^2 + b^4) ≥ 3(ba^3 + ab^3)
2[(a^2 + b^2)^2 - (ab)^2] ≥ 3ab(a^2 + b^2)
2[((a+b)^2 - 2ab)^2 - (ab)^2] ≥ 3ab((a+b)^2 - 2ab)
2((a+b)^2 - 2ab)^2 - 2(ab)^2 ≥ 3ab(a+b)^2 - 6(ab)^2
2((a+b)^4 - 4ab(a+b)^2 + 4(ab)^2) - 2(ab)^2 ≥ 3ab(a+b)^2 - 6(ab)^2
2(a+b)^4 - 11ab(a+b)^2 + 12(ab)^2 ≥ 0
(2(a+b)^2 - 3ab).((a+b)^2 - 4ab) ≥ 0 [*]
(a+b)^2 - 4ab
= a^2 + 2ab + b^2 - 4ab
= a^2 - 2ab + b^2
= (a-b)^2
≥ 0
2(a+b)^2 - 3ab
= 2(a^2 + 2ab + b^2) - 3ab
= 2(a^2 + b^2) + ab
Let a=r.cos(t) and b=r.sin(t):
= 2r^2(cos^2(t) + sin^2(t)) + r^2.sin(t).cos(t)
= r^2(2 + sin(2t)/2)
≥ r^2(2 - 1/2)
= (3/2)r^2
≥ 0
So the 2 terms on the LHS of [*] are indeed non-negative.
Thanks for solving and explaining this math problem. I was really stuck on it. Turns out that you're not supposed to replace a^2+b^2 with m and ab with n as I did.
Came for math, subbed for soothing voice
❗❗❗Lithuania mentioned ❗❗❗
Great video as always!
The slide in the beginning has 3 in the denominator on the right hand side instead of 2
That was my tired brain. I fixed it. Thanks
Nice videos, great explanation. I hope that you will upload a new video about ramanujan nested radicals especially the formula..😁👍
Thank you for the nice problem and its solution. I solved it in a different way, thus maybe worth sharing: If one of a and b is equal to 0, then the RHS of the inequality is 0 and the inequality is true, so we can from now on assume that a*b 0. If one of a and b is negative, than the RHS is either negative and the inequality is true or the other value is also negative and the RHS is positive and this is equivalent to the case where both a and b are positive, and we cannot say at this point whether the inequality is true, so it suffices to show that the inequality is true for a and b both positive. Under this assumption one can divide both sides of the inequality by a^2*b^2 and obtain the equivalent inequality ((a/b)^2+1+(b/a)^2)/3 = ((a/b+b/a)^2-1)/3 >= (a/b+b/a)/2, which after moving the RHS to the LHS is equivalent to ((a/b+b/a-3/4)^2 - 25/16)/3 >= 0. At this place it is very useful to remember that due to AM-GM one has that a/b+b/a >= 2, which leads to ((a/b+b/a-3/4)^2 - 25/16)/3 >= ((2-3/4)^2 - 25/16)/3 = ((5/4)^2 - 25/16)/3 = 0, the inequality following from the fact that the square function is increasing on [0, \infty), which completes the proof.
I like your solution a lot more, but think my solution is correct and solves in a different way (and doesn't assume knowledge of AM-GM inequality) by looking at different domains of a, b. Note that left hand side is always greater than or equal to zero due to even exponents. Right hand side is greater than or equal to zero if a,b are both positive, or both negative otherwise the solution is less than or equal to zero and inequality is trivially true. So we only need to focus on the two cases
1) a >= 0 and b >= 0
without loss of generality assume a >= b and rewrite at a = b c where c>= 1 then substitute and solve in terms of b. After collecting terms you get the inequality
b^4 (c^4 + c^2 + 1)/3 >= b^4 (c^3 +c)/2
clearly this inequality holds for all c>= 1 due to quartic power
2) a < 0 and b < 0
without loss of generality assume a = 1 then substitute and solve in terms of b. After collecting terms you get the same inequality as 1
We can manipulate the given inequality into a sort of sum of squares form. First, notice that we only need to prove the inequality for positive a and b, since when a and/or b are negative, the LHS remains the same but the RHS can only decrease or remain the same. Now, for simplicity, we can get rid of the denominators by multiplying the inequality by 6 and move all terms to LHS yielding:
2a^4 + 2b^4 + 2a^2 b^2 - 3a^3 b - 3a b^3 >= 0
This, in turn, can be represented as a (sort of) sum of squares and fourth powers as:
(a - b)^4 + (a^2 - b^2)^2 + ab (a-b)^2 >= 0
The "sort of" is due to the fact that there is ab in front of the last square, but since we already justified focusing on non-negative values of a and b in the proof, we are good to go.
I solved by using direct AM-GM inequality. It worked just fine because a^4,b^4 and a^2b^2 are non-negative so you can apply the theorem.
Great video!
The proof seems almost immediate to me: first we increase 2ab with a^2+b^2, obtaining on the second member (a^2+b^2)^2/4
Then the given inequality with the second member increased is easily verified (a perfect square greater than or equal to zero is obtained, clearly always true).
Thanks Sir
I tried this myself. I tried substituting t = a²+b² and u = ab and solving resulting second order equations for t and u, but it led nowhere. I thought the trick is to show a and b must be real (not complex), so I hoped for the discriminates to help but no such luck. Seeing your solution illustrates how off I was in my attempt.
Tried a similar method. But instead of finding discriminant, I factored out a polynomial 2t^2-2u^2-3tu to get (2t + u)(t-2u)>=0. Knowing that t>=0 and t>=u (from a^2+b^2 = 2ab) it still led me nowhere.
@@Lightseeker1-j5pif you divide by a²b² and note x=a/b you'll obtain a nice inequality where X=x+1/x appears.
At the end you have to solve 2X²-3X-2>=0.
I put everything on the same side, and it looks like (a-b)^4, then I keep simplifying the inequation until I got to (a-b)²(2a²+ab+2b²) >= 0. From here we get to a²+b² >-= -ab/2, which I think is always true, but I don't know to explain why.
Makes sense. You just need to show that it is always true.
For the first time in many videos I could not understand your demonstration since I could not figure how the first equality of the proof came from. I guess I have to study more algebra. Thanks anyway.
a²+b² = (a-b)² +2ab
@@PrimeNewtons Thank you for your reply but my question is how a²b²=2a²b²-a²b² since a^4 and b^4 can be cancelled?
Here's a math question:
Solve for x:
x^3 + 2x^2 - 7x - 12 = 0
Please solve this equation I will be thankful
Can you please clarify as to what you did at 5:21?
The arithmetic-geometric mean inequality you cited as justification for some of your steps is true if a and b are positive. How is your proof valid for all real a and b?
This is because a and b are real numbers and the square of any real numbers are positive always. That's why the property AM>=GM holds for a^2 and b^2 as both are greater than or equal to zero. Hope this helps.
I don't agree. You use the same letters a and b for different values. Our initial a and b are not necessary positive numbers. a2 and b2 are posirive, but a and b not necessary. So, I don't see that you don't need to use the absolute value.
Anywere, I think it don't affect the result because if one of then is negative, the inequality is more obvious because a3 or b3 will be negative numbers in the right expression and it will be minor.
In math, you should not agree or disagree. Just be right.
I mean you use the same letters to enunciate that the arithmetic mean is greather than geometric mean. You should not use a and b for this enunciate, because here a and b are positive numbers, but out original a and b are not. I think you should enunciate the theoreme with another letters. The a and b letters of theoreme are positive numbers, but our initial a and b are not. It's true that a2 and b2 are positive and you can use the theoreme, but when you cancel the square root I think you may use absolute value.
I like your videos and I watch all or them. My comments are only with constructive intentions. Sorry for my english.
@@user-ky9kv5je9s I understand what you mean. Thank you!
(a^2+b^2)/2 >= ab because (a-b)^2 >= 0.
Why not just develop (a+b)^4 ?
tried it - did not work
Two questions:
1. You say that a, b > 0 when employing the AM-GM inequality, yet according to the question, a and b can be negative.
2. At 5:37, you squared the RHS but not the LHS.
It was a mistake to use a,b. Assume I used u,v.
@@PrimeNewtons If you mean u = a^2 and v = b^2, it renders all of your subsequent "proof" meaningless.
And why are you assuming positive values for a, b, when they can be negative? The AM-GM inequality doesn't hold for negative values.
@@Straight_Talk I thought my comment would help you. It didn't. And I agree.
Bro i am from india i have a challenge bro please solve this if (a+1)(b+1)(c+1)(d+1)=1 and (a+2)(b+2)(c+2)(d+2) =2 and (a+3)(b+3)(c+3)(d+3) =3 and (a+4)(b+4)(c+4)(d+4)= 4 then find (a+5)(b+5)(c+5)(d+5) Btw a i am also mathematics lover
Interestingly,
(a⁴ + a²b² + b⁴)/3 >= a²b² by the AGM 😊.