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Here is a problem to start with :
Let be real numbers such that the polynomial equation
has eight positive real roots. Determine all possible values of
Easy Math Editor
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Top NewestSince we were given that the roots were positive reals,we get motivated and tempted to use,R.M.S≥A.M≥G.M≥H.M,also since we have been given,∑α and ∑αβ we try to find the value of ∑α2=42−2×7=2.We apply R.M.S≥A.M 82≥84⟹21≥21,hence equality occurs,meaning that all the roots are equal.Now,that means,A.M=G.M,hence84=8f⟹2561=f,hence there is only one value of f=2561.Is this correct Harsh?
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In your solution you need prove that (cyclic∑ω)2−cyclic∑ω1ω2=cyclic∑ω2
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That's an identity that is true for all numbers, real and imaginary. And there should be a 2 infront of that second summation.
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If x,y,z are positive real numbers, prove that,
(x+y+z)2(yz+zx+xy)2≤3(y2+yz+z2)(z2+zx+x2)(x2+xy+y2)
Give me as much different solutions please .
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U may use the concept of fermat pt to solve this one..then it will be very easy...suppose three line segments AF,BF,CF of length x,y,z meet each other at120 degrees and now use cosine rule to find out the lengths of the triangle and see that each term on the rhs of the inequality is actually squares of the sides...then express xy+yz+zx as 34T where T is the area of the triangle....then use T=4Rabc and see that the inequality reduces to 3R greater than or equal to (x+y+z) which is obvious since F is the fermat pt
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A very elegant solution
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You have a set of consecutive positive integers,{1,2,3,...,199,200},if you choose 101 numbers from these at random,then prove that there at-least two numbers of which one divides the other.
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If x and y are integers such that y2+3(xy)2=30x2+517, Prove that 3(xy)2=588
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Let y2=b and x2=a.
b(1+3a)=30a+517
⟹b(1+3a)−10(3a+1)=507
⟹(b−10)(3a+1)=507×1=13×39=3×169
Since a , b are integers, after checking through all 6 cases, we get only admissible values of (x,y)=(±2,±7).
This forces 3(xy)2=588.
\QED
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Find all positive integer solutions a, b and c such that
a27+b211=77c.
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THIS IS ONLY AN OUTLINE OF THE SOLUTION, NOT A COMPLETE SOLUTION.
a27+b211=77c.
Case 1: c>1
This forces a2<7 and b2<11
Now it is easy to check that (a,b,c)=(1,1,1386) and (a,b,c)=(3,3,154) satisfy the given conditions.
Case 2: c≤77
Since c≤77 this implies 77c≤1.
Let 77c=k1 for some integer k≥1.
⟹c=k77
Now since c is a positive integer , this forces k | 77.
Possible values for k=1,7,11,77.
Now plugging the possible values of c in the given equation and using Simon's Favourite Factoring Trick, we can easily solve for a,b
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Find all primes p such that for all prime q, the remainder of p upon divison by q is squarefree.
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I think that 3 is the only prime number. I used the fact that all primes>3 is of the form ±1(mod6).
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x, y, z are positive real numbers such that x^2+y^2+z^2=3 Prove that x+y+z >= (xy)^2+(yz)^2+(zx)^2
Sorry for not using latex as I was in a hurry.
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Can u send the solution of this question ? I have been trying. No progress ....
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I didn't get it.
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Let x,y and z be positive real numbers such that x+y+z=1. Prove that 1+xyz+1+yzx+1+zxy≤41.
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I got this. This is simple AM-HM inequality. Write 1+x=(1-y)+(1-z). Similarly for others
Apply AM-HM and then manipulate on the RHS.
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Find all reals x,y,z in (1,∞) such that
x+y+z+x−13+y−13+z−13=2(x+2+y+2+z+2)
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https://brilliant.org/problems/three-variables-one-equation/?group=4bQcwX9pLM2R#!/solution-comments/106741/
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If a,b,c are positive integers such that a1+b1+c1<1. Prove that the maximum value of a1+b1+c1 is 4241
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If none of a,b,c=2, the maximum value of the expression will be 31+41+51=6047<4241. So, one of them has to 2 say a. So we have to prove that the maximum value of b1+c1 is 2110. If none of b,c=3, the maximum of the expression will be 41+51=209<2110. So, one them has to be 3 say b. Therefore, c1≤71.Therefore the maximum value of c is 7 and the maximum value of a1+b1+c1 is 4241.
Not sure if this is the best method.
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Find all primes x,y such that xy−yx=xy2−19.Enjoy solving this.
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Using FLT, we have
xy≡x(mody)⇒x≡−19(mody)⇒x+19≡0(mody)
yx≡y(modx)⇒−y≡−19(modx)⇒19−y≡0(modx)
From here, it is clear that either x=2 or y=2. We now separate this into 2 cases:
Case 1: y=2
When y=2, we have 19−2=17≡0(modx), which implies x=17. The solution in this case is (17,2). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.
Case 2: x=2
When x=2, we have 19+2=21≡0(mody), which implies y=3 or y=7. Checking, we find both these solutions work. Thus, 2 solutions exist in this case: (2,3) and (2,7).
Therefore, only 2 solutions exist: (2,3) and (2,7).
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would you please explain how x = -19 mody
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xy−yx=xy2−19 in (mody), we get x=−19(mody).
PuttingLog in to reply
Using FLT and modulo x and y we get that,x+19≡0(mody),y−19≡0(modx),now just use parity(as suggested by @Sharky Kesa).
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Let p and q be distinct primes. Show that
j=1∑(p−1)/2[pqj]+j=1∑(q−1)/2[qpj]=4(p−1)(q−1).
Note that [ ] represents the greatest integer function.
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Let S=(x,y)∣x,y∈N,1≤x≤(p−1)/2,1≤y≤(q−1)/2.
We note that the set S can be partitioned into two sets S1 and S2 according as qx>py or qx<py. Note that there are no pairs in such S such that qx=py.
The set S1 can be described as the set of all pairs (x,y) satisfying 1≤x≤(p−1)/2,1≤y<qx/p. Then S1 has j=1∑(p−1)/2pqj elements.
Similarly, S2 has i=1∑(q−1)/2qpi. Hence, we get the required result.
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Let a,b,c,d∈N such that a≥b≥c≥d and d=0. Show that the equation x4−ax3−bx2−cx−d=0 has no integer solution in x.
Please post a "complete solution" like you would do in RMO.
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I'll have a crack at it. As d=0, x=0.
Denote y=x1.
We have
x4=ax3+bx2+cx+d
1=ay+by2+cy3+dy4(1)
Now, we separate this into 2 cases:
Case 1:
x=−n where n∈N, then y=−n1. Substituting this into (1), we get
n−a+n2b+n3−c+n4d=1
n2−an+b+n4−cn+d=1
Here, the LHS is non-positive since a≥b and c≥d, whereas the RHS is positive. Thus no solutions exist in this case.
Case 2:
x=n where n∈N, then y=n1. Substituting this into (1), we get
na+n2b+n3c+n4d=1
This implies that n=1, and
na<1
1≤na+n2a+n3a+n4a<i=1∑∞nia=n−1a
These two equations imply that
an>1
an−1<1
There are no such n that satisfy the above two inequalities. Thus, no solutions exist.
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Isn't this from INMO 2013 ??
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Yes it is.
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let n be an integral solution (if possible). then n [n^3-an^2-bn-c] = d. Now for the equation to hold n divides d. so n is less than or equal to d. If n is negative -ax^3 is positive and is greater than bx^2 as a is greater than b.similarly for the next term and the total value becomes positive but known the value is 0.hence our assumption that n is negative is wrong .so n is positive and less than d. Now as n<d <a, n^4<an^3 and -bn^2 and -cx and -d all are negative,so the sum is also negative which was again 0. hence contradiction. So no integral roots exists.
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Let a,b,c be positive reals satisfying (a+b)(b+c)(c+a)=1.
Prove that ab+bc+ca≤43
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Using am gm
(a + b) + (b + c) + (c + a) > 3
a + b + c > 3/2
now using cauchy,
3(a2+b2+c2)>(a+b+c)2
so
a2+b2+c2>3/4
using cauchy again,
(a2+b2+c2)2>(ab+bc+ca)2
3/4 > ab + bc + ca
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Nice solution. You could also have used the fact that (a+b+c)2≥3(ab+bc+ca)
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Dev,your solution is incorrect. Check the solution again,you will find your flaw.
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You got the following two inequalities,a2+b2+c2>ab+bc+caanda2+b2+c2>33,what next?How can you conclude that,43>ab+bc+ca?
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(sum(a^2))(sum(b^2)) > (sum(ab))^2
sum is cyclic
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If a,b≥0. Prove that 2(a(a+b)3+ba2+b2)≤3(a2+b2).
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Prove that 2<(n+1n+2)(n+1) for all integers n>0.
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Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nx
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Let p>3 be a prime number. Suppose k=1∑p−1k1=ba where gcd(a,b)=1. Prove that a is divisible by p2.
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I am able to prove that a is divisible by p. But not able to prove divisible by p^2. Could u send the solution ?
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Please post the proof, thanks!
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I have the solution only till a is divisible by p. To prove that a is divisible by p2 is given below the solution as a challenge.
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In a triangle ABC , D is the midpoint of side AB and E is the point of trisection of side BC nearer to C.
Given that ∠ADC=∠BAE, find ∠BAC.
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Is the answer angle BAC = 90 degrees ? I hope so My proof : Apply Menelaus theorem on triangle BCD with transversal EA. Obtain that F is the midpoint of CD. Then its simple angle chasing :)
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Shrihari,we get that CF=2DF
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Yup I also got 90 degrees.
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Yes , Menelaus is indeed a good approach for this problem. For shortening the solution: Once you get F is the midpoint of CD , you can directly say that ∠BAC=90∘ , do you see why?
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There are given 2500 points on a circle labeled 1,2,...,2500in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.
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Given seven arbitrary distinct real numbers, show that there exist two numbers x and y such thst 0<1+xyx−y<31.
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The question is easy. Consider x and y to be some tan(A) and tan(B) respectively and consider A and B to vary from 0 to pi so that all reals are covered. now divide it into six equal portions spaced by pi/6. Now u have 6 pigeons and 7 holes and you are done. Then that expression is simply tan(A-B).
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If a circle goes through point A of a parallelogram ABCD, cuts the two sides AB,AD and the diagonal AC at points P,R and Q respectively. Prove that (AP×AB)+(AR×AD)=AQ×AC.
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I hope you all have drawn the diagram.(Consider all the segments below as vectors in the given direction.)
Let AS be the diameter of given circle.
AB+AD=AC
Therefore taking dot product withAS
AB.AS+AD.AS=AC.AS
Using, a.b=∣a∣∗∣b∣∗cos(θ) [θ is angle between the vectors]
∣AB∣∗∣AP∣+∣AR∣∗∣AD∣=∣AQ∣∗∣AC∣
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Let ABC be a triangle and P be a point inside it.Let x,y,z denote the lengths of PA,PB,PC.Let a,b,c be the lengths of sides BC,CA,AB.Find all points P such that axy+byz+czx=abc.
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Find the least positive integer n such that 2549 divides n2545−2541
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Please post the solution , Thanks!
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Let ABC be an acute-angled tiangle .Let O denote its circumcenter.Let Π be the circle passing through the points A,O,B.Lines CA and CB meet Π again at P and Q respectively.Prove that PQ is perpendicular to the line CO.
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Harsh,could you please add a figure?
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Hint: Let QP∩CO={R} and CO∩Π={S} and then it will suffice to prove that ΔCRP∼ΔCAS
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An easy INMO geometry problem:
Let ABC be a triangle with circumcircle Γ. Let M be a point in the interior of triangle ABC which is also on the bisector of ∠A. Let AM,BM,CM meet Γ in A1,B1,C1 respectively. Suppose P is the point of intersection of A1C1 with AB and Q is the point of intersection of A1B1 with AC. Prove that PQ is parallel to BC .
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Let a,b∈R,a=0. Show that a2+b2+a21+ab≥3.
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Well the question becomes very easy if u make it a quadratic in b and then prove that the Discriminant is less than 0. I did it that way.
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If a,b,c,d,e are real numbers, prove that the roots of x5+ax4+bx3+cx2+dx+e=0 cannot all be real if 2a2<5b.
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Let roots be p,q,r,s,t.
Sum of roots = -a & Product of roots taken two at a time = b.
Therefore Sum of squares of roots = a2−2b
If all roots are real, then by Titu's Lemma,
a2−2b≥5a2
⟹2a2≥5b
Therefore the polynomial cannot have all roots real if 5b>2a2.
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Show that n5+n4+1 is not prime for n>1.
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n2+n+1 is factor of the given expression.Hence the result is obvious.
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Prove that
1<10011+10021+.........+30011<34
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It is easy. I got a calculus and a non calculus answer
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Since Rmo is a precalculus olympiad, answer without calculus will be better:P
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this is a copy paste from this note on moderators
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If m is a positive real number and satisfies the equation 2m=1+m1+m2−41 Find m.
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Could you send the solution of this question. I tried expanding and landed up with a fourth degree polynomial whose roots I am unable to guess
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Find all pairs (m,n) such that 2m+3n is a perfect square. (A popular NT problem)
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Its easy.Only one solution (2,2)
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Wrong. Check again.
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gcd(2m,3n)=1, we use the general formula for pythagorean triplets. Taking various cases,we get that only m=4,n=2 are the solutions.
Taking mod 4 we get that m,n should be even. Then usingLog in to reply
Let d(k) denote the sum of the digits of k in base10. Find all natural numbers n such that n, d(n), d(d(n)) sum to 2015.
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I am getting no solution. Is that right ? My proof : n,d(n),d(d(n)) are congruent to the same number say 'x' (mod 9). So their sum is congruent to 3x (mod 9). Therefore 2015 is congruent to 3x ( mod 9) which is impossible as 3 does not divide 2015
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Here is a geometry question. Waiting for numerous solutions. In a triangle ABC, excircle opposite to A touches BC at point Q. Let M be the midpoint of BC. Incircle of triangle ABC touches BC at point P. Prove that PM=MQ. I have been trying this question but I am stuck. So please solve this question.
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Its easy , note that BP=QC=s−b where s is the semiperimeter of the triangle and b is length of side opposite to ∠B respectively. Since M is mid point of BC , the result follows.
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Are protractors allowed for RMO? Nothing about it is mentioned in the website and the hall ticket.
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Oh nice !
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Try this set guys.
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1991k∣(199019911992+199219911990)
max(k)=?
Details:
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1991
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How ? Could u give some rough outline of your solution ?
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https://brilliant.org/problems/triangular-grid-problem/?group=U3Rx70QSpE0z&ref_id=1044906
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Yes..... Can you post solution ?
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Everyone here who have posted questions on this board and have not got any solutions please post the solutions now as RMO is only 42 hours from now. Its better that we understand the methods. Sorry that I cannot send the solution of my geometry question as I myself have been trying that question but could not arrive at the solution.
And please post some more questions if possible :)
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Done :)
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Thanks
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Let x,y be real numbers such that x,y>1
Prove that y−1x2+x−1y2≥8
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Just use AM-GM inequality directly on the expression in the LHS and then use the fact that (a-2)^2>0 Then rearrange a bit. You get the required answer
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Alternatively we can substitute p = x-1 and q = y-1.
Then it will be easy manipulate the expression.
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Nice idea!
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Since the expression is symmetric, let's assume without loss of generality that x≥y.
y−1x2+x−1y2≥y−1y2+x−1y2≥4+x−1y2⟹y−1x2≥4. Minimum value of y−1x2 is 4 and it occurs if and only if x=y=2.
Therefore, y−1x2+x−1y2≥8.
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Hmm you have not proven that min value of y−1x2 is 4.
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y−1y2 is 4?
Do you mean that I haven't proven that minimum value ofLog in to reply
Sorry didn't see this post but since I found it, here you go - 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.
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a,b,c are real numbers such that a+b+c=0 and a2+b2+c2=1. Prove that a2b2c2≤541.
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See my note
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A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw.
What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, is repeated?
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1/2
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Correct. Can you show your steps?
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Write the sum Sn=k=0∑nk3+9k2+26k+24(−1)k(kn) in the form q(n)p(n) where p(n),q(n) are polynomials with integer coefficients.
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We first note that k3+9k2+26k+