I didn't knew there were multiple winning strategies for ##P##!
Also, couldn't we generalise that further for when ##P## chooses any value of ##c## in the first move, then the condition for $$f(1)<0$$ would change to $$a+b+c+1<0$$
So if ##Q## chooses a value of ##a## then ##P## would choose...
When I first used ##A.M \geq G.M## on$$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$
the equality would never hold because if $$c=c-2x=c-2y=c-2z$$
then we would get $$x=y=z=0$$ which was not possible
So, basically as I understand it,
we had $$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$
And ##c## is a constant, so we use ##A.M \geq G.M## on $$g(x,y,z)=(c-2x)(c-2y)(c-2z)$$
So, we get,
\begin{align*}
\dfrac{(c-2x)+(c-2y)+(c-2z)}{3} &\geq \left((c-2x)(c-2y)(c-2z)\right)^\frac1 3\\
\dfrac{3c-2c}{3} &\geq...
Are you saying that I must prove why the equality holds? or do you want me to show when the equality holds?
As I said that for ##a_1,a_2, ... ,a_n##, ##A.M=G.M## when ##a_1=a_2= ... =a_n##
So in our case the equality should hold if $$x+y+z=x+y-z=x-y+z=y+z-x$$
But this equality will only hold...
I think I have just a strategy but I am not sure because lets say that P plays according to my strategy and chooses a negative value of ##b## say ##b=-1## but then after Q chose a value of ##c## then I don't know what value of ##a## P should choose, I know (more like I feel) that there are some...
I just noticed that $$c(c-2x)(c-2y)(c-2z)$$
Looks similar to the formula for area of a triangle!
Is that what you meant, is geometrical meaning of ##f(x,y,z)## is that is 16 times the square of the area of a triangle with sides ##x,y,z##! because area of a triangle with sides ##x,y,z## will be...
I don't know about why is that maximum, if equality holds then they should be maximum. I think I don't understand what you're trying to say?
Also even if ##x,y,z \gt 0##, ##x+y-z## can still be less than zero and we cannot use A.M G.M inequality the way I did, if ##x,y,z## were sides of a...
I might be wrong, but I thought that when ##a=0## then, if the value of ##b## is chosen to be positive, then the equation ##f(x)=x^3+ax^2+bx+c=0## should have only one solution because we can see that
##f'(x)=3x^2+2ax+b##
Now to have two different values of ##x## where ##f'(x)=0## we should have...
But I applied ##A.M \geq G.M## on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$
##(z-x+y),(z+x-y),(x+y-z)## and ##(x+y+z)## aren't necessarily positive right?
Also to answer your questions in post #12, I have no idea about the geometric meaning of ##A.M \geq G.M##
All I knew was that if we had ##n##...
Do you want me to derive the equation of director circle?
Let ##y=mx+c## be a tangent to the ellipse ##\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1##, so on solving these two we get,
\begin{align*}
b^2x^2+a^2(mx+c)^2&=a^2b^2\\
(b^2+a^2m^2)x^2+(2a^2mc)x+a^2(c^2-b^2)&=0
\end{align*}
For ##y=mx+c## to be a...
I did realise my mistake, thank you for your help and also for showing how to work backwards from ##\frac{\binom{2n}{n}}{2^{2n}} < \frac{1}{\sqrt{2n+1}} ## to ##
\frac{\binom{2n+1}{n}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}}##
Now this looks quiet confusing to read as a third person who doesn't know whats going in my head, but to me it makes perfect sense!
Still if someone wants to read that, then the case ##e<f## was in the original attempt (see post #21 of this thread) ##\dfrac{\binom {2n} n}{2^{2n}}<...
I think I figured out what was wrong
(refer to my first attempt of problem 15)
I wanted to show that $$b<c$$
I knew that $$e<f$$
is true.
I assumed that ##b<c## is also true
First I showed that ##a<b## and then I showed ##c<d##
so I concluded that $$a<d$$
Now I did the same process with...
On rereading, I get this feeling that may be I'm doing something wrong here, but I'm confused as we are dealing with inequalities here, if we had been dealing with equalities then everything you said makes sense to me.
I do agree with what you said here, I also agree with your solution which was discussed earlier in this thread, but I think that there is another method of doing this problem which is with the use of orders ##a<b<c<d## which I mentioned above.
But this statement is confusing for me.
If the...
Hey @julian, Thanks for your reply, but I still have the same question,
you said that,
\begin{align*}
\frac{\binom{2n+1}{n}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}} \qquad (*)
\end{align*}
is correct
But as we have,
\begin{align*}
\binom{2n}{n} &< \binom{2n+1}{n}\\
\dfrac{\binom {2n}...
Problem 12
Let ##\sqrt{n}+\sqrt{n+4}=p## such that ##p\in\mathbb{Q}##
So we get,
$$\begin{align}
\sqrt{n}+\sqrt{n+4}&=p\nonumber\\
n+n+4+2\sqrt{n}\cdot\sqrt{n+4}&=p^2\nonumber\\
\sqrt{n}\cdot\sqrt{n+4}=\frac{p^2-2n-4}{2}&=q\space\text{(say)}\nonumber
\end{align}$$
We can see that...
I shouldn't write ##c^2=(k-2)(k+2)## I should have directly wrote that from ##c^2+2^2=k^2## the only solution we get is ##c=0## and ##k=2## (as there are no pythagorean triplets with 2), I was thinking something else when I wrote ##c^2=(k-2)(k+2)##.
I don't like that solution anyway, maybe I...
@julian can I ask what was the mistake in this attempt (Its not particularly clear to me)?
I think that according to you until here,
$$\dfrac{\binom {2n+1} {n}}{2^{2n+1}}< \dfrac{1}{\sqrt{2n+3}}$$
the above expression is correct, but after that why can't we write,
$$\dfrac{\binom {2n}...
\begin{align*}
\frac{1}{2} \cdot \frac{2n+1}{n+1} < \sqrt{\frac{2n+1}{2n+3}}
\end{align*}
Now, since both sides are positive, we can square both sides,
\begin{align*}
\frac{1}{4} \cdot \frac{(2n+1)^2}{(n+1)^2} &< {\frac{2n+1}{2n+3}}\\
\frac{1}{4} \cdot \frac{(2n+1)^2}{(n+1)^2} -...
My thinking behind writing "for smallest solution... equality holds" was that, suppose we are given
$$x \leq y$$
And we have to find the minimum possible value of ##y## for which the above relation is true for all possible values of ##x## (both ##x,y\in \mathbb{R}## are variables)
Now clearly in...