@@tianqilong8366 There is a general formula for quartic, but i didn't use it here. Instead you can observe that this particular equation has a nice , symmetric form. You can divide by k^2 and consider q=x^2 + (1/x^2). It is called reciprocal equation
Simplify by repeated substitutions: Divide through by y^4, write t=x/y, and expand numerator, (1 + t + t^3 +t^4) / (1 + t^2)^2 Divide through by t^2 and regroup, (1/t^2 + t^2 + 1/t + t) / (1/t + t)^2 t>0 implies w = 1/t + t ≥ 2. Rewrite and complete the square, (w^2 - 2 + w) / w^2 = 1 - 2/w^2 + 1/w = -2(1/w -1/4)^2 + 9/8 which has a maximum of 9/8 at w=4 and a minimum of 1 at w=2. Max occurs when t^2-4t+1=0, t=x/y=2±√3. Min when t=x/y=1. Also shows that original expression can be written as a quadratic in 1/w, ie, in xy/(x^2+y^2)
For the second inequality, use polar coordinates. Since both x and y are positive, we can assume θ is between zero and π/2. The r’s will cancel, and the remaining expression can be written as a quadratic function of sin(2θ), which has a maximum value of of 9/8 on the domain of θ. At least, I think so. I haven’t put pencil to paper yet. I’ll leave it as an exercise to the reader 😆
@@sr7821 thanks for checking! I like trying to solve some of these problems mentally on long road trips. Test the capacity of how much I can keep up with in my head.
I don't understand why at 3:20 it is that the square of -4xy is 18x²y². Shouldnt it be 16x²y², so we add to the square trinomial 2x²y², which does not change the final outcome
From the final result, since x and y are symmetric, if x/y = 2+sqrt(3), then y/x = 2-sqrt(3), implying 2+sqrt(3) = 1/(2-sqrt(3)). It doesn't take much to verify this, but it is pretty similar to the result that the roots of x^2-x-1 are phi and -1/phi.
@Suman Gupta For this particular problem it might not be important. But seeing things from different points of view and generalizing or being generalized from another result i super important. A bit lightheartedly one could say it's always the "next" problem that matter. It's also fine to be able to concentrate on the problem at hand of course. It's more important to be always learning than to be always right as you know.
Any chance to talk about IMO 2022 problems? Would love to see more IMO problems (2000 onwards)
You can consider y = kx, with k>0. The given expression can be written as 1 =0
The second one will be: 8[(1+k)^2][1-k+k^2]^2 - 9[1+k^2]^2
Looks cool, but how do you know the exact answers for quartic equation? Is there any formula for that?
@@tianqilong8366 There is a general formula for quartic, but i didn't use it here. Instead you can observe that this particular equation has a nice , symmetric form. You can divide by k^2 and consider q=x^2 + (1/x^2). It is called reciprocal equation
@@eduardciuca217 thanks for the calrification. Its cool!
Simplify by repeated substitutions: Divide through by y^4, write t=x/y, and expand numerator,
(1 + t + t^3 +t^4) / (1 + t^2)^2
Divide through by t^2 and regroup,
(1/t^2 + t^2 + 1/t + t) / (1/t + t)^2
t>0 implies w = 1/t + t ≥ 2. Rewrite and complete the square,
(w^2 - 2 + w) / w^2 = 1 - 2/w^2 + 1/w = -2(1/w -1/4)^2 + 9/8
which has a maximum of 9/8 at w=4 and a minimum of 1 at w=2. Max occurs when t^2-4t+1=0, t=x/y=2±√3. Min when t=x/y=1. Also shows that original expression can be written as a quadratic in 1/w, ie, in xy/(x^2+y^2)
Really great approach! Totally different than anything I thought of. Incidentally, your 1/w is the same as 1/2*sin(2θ) from Davey Jones' solution.
For the second inequality, use polar coordinates. Since both x and y are positive, we can assume θ is between zero and π/2. The r’s will cancel, and the remaining expression can be written as a quadratic function of sin(2θ), which has a maximum value of of 9/8 on the domain of θ.
At least, I think so. I haven’t put pencil to paper yet. I’ll leave it as an exercise to the reader 😆
yeah, it does reduce to a quadratic in sin(2θ). Nice!
@@sr7821 thanks for checking! I like trying to solve some of these problems mentally on long road trips. Test the capacity of how much I can keep up with in my head.
@@DaveyJonesLocka So do i. But i try solving it mentally cuz I'm too lazy to pick up a pen 😂
I don't understand why at 3:20 it is that the square of -4xy is 18x²y². Shouldnt it be 16x²y², so we add to the square trinomial 2x²y², which does not change the final outcome
From the final result, since x and y are symmetric, if x/y = 2+sqrt(3), then y/x = 2-sqrt(3), implying 2+sqrt(3) = 1/(2-sqrt(3)). It doesn't take much to verify this, but it is pretty similar to the result that the roots of x^2-x-1 are phi and -1/phi.
THankS SO MuCH111 lOVe yOUr VidEOs!
Nice problem, though somehow the purely algebraic solutions left me unsatisfied. Why is 1 the minimum, why is 9/8 the maximum, and when do they occur?
Where is the problem from?
It’s from Turkey Junior National Olympiad 2011
What was the meaning of taking x+y=a and xy=b when u have ALREADY SOLVED it in terms of x and y before???
it's just another way to solve bro
@@gdtargetvn2418 yes it's an aliter
@Suman Gupta For this particular problem it might not be important. But seeing things from different points of view and generalizing or being generalized from another result i super important. A bit lightheartedly one could say it's always the "next" problem that matter. It's also fine to be able to concentrate on the problem at hand of course. It's more important to be always learning than to be always right as you know.
The first one Cauchy's inequality
I noticed that too, you are right
You mean convert x+y to (x^0.5)^2 + (y^0.5)^2 and do it again on x^3+y^3 to prove >=1 ?
@@tianqilong8366 yes
It would be more logical to prove the inequalities by contradiction,
assuming the converse and showing it to be false.