Let x=a+b, y=a+c, z=b+c.
nice method, mine was a little longer:Let x=a+b, y=a+c, z=b+c.
Then LHS=[(x+y-z)/z+(x+z-y)/y+(y+z-x)/x]/2=[(x/y+y/x)+(x/z+z/x)+(y/z+z/y)-3]/2 >= 3/2.
(We used AM-GM thrice in the last inequality to bound the (t+1/t) terms below by 2.)
that seems familiar to a question in the maths competition by AMT last year. couldnt do it lolsuppose that a polynomial with integer coefficients satisfies P(100)=100. Find the maximum of integer solutions to P(k)=k^3 such a polynomial can have.
Can you please clarify the exact question statement?suppose that a polynomial with integer coefficients satisfies P(100)=100. Find the maximum of integer solutions to P(k)=k^3 such a polynomial can have.
a polynomial is called self centered if it has integer coefficients and p (100) = 100. if p (x) is a self centred polynomial, what is the maximum number of integer solutions to p (k) =k^3?Can you please clarify the exact question statement?
Are these polynomials definitely required to have integer coefficients?
Is it definitely P(100)=100?
Are we meant to find the maximum number of solutions to f(k) = k^3 possible, or the largest possible k such that f(k)=k^3?
(I just want to be clear the problem is correctly written before spending time on it.)
is the answera polynomial is called self centered if it has integer coefficients and p (100) = 100. if p (x) is a self centred polynomial, what is the maximum number of integer solutions to p (k) =k^3?
nope, I think you've taken a wrong approachis the answer
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EDIT: n-2
Nesbitt's inequality is nice.
Cauchy-Schwartz states for real variables we have:Nesbitt's inequality is nice.
Should I post a hint?here is a great question I was doing a couple of weeks back. show that in every set of 9 positive integers, a subset of 5 positive integers exists such that the sum of the numbers is divisble by 5.
Or is it the wording of the question that is throwing people off? Here is the exact wording of the question.Should I post a hint?
Cauchy-Schwartz states for real variables we have:
(ax+by+cz)^2 (a^2+b^2+c^2)(x^2+y^2+z^2)
with equality iff (x,y,z)=(ta,tb,tc) for some real c.
Lots of proofs of Cauchy-Schwartz are on this site, so I will assume it. (Can provide a short proof if you like, such as the one based on non-negative polynomials having non-positive discriminant).
Then C-S implies
LHS^2 =< ((6x+1)+(6y+1)+(6z+1))(1+1+1)=27.
And taking square roots of both sides finishes the question. The C-S equality condition also implies x=y=z=1/3 if we have equality.
You have some inequalities the wrong way around, idk if your method works exactly but some tweaking might fix it.
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Yes I will fix it give me a moment.You have some inequalities the wrong way around, idk if your method works exactly but some tweaking might fix it.