$\textbf{Problem 01}$
Let $ x,y,z\in \mathbb{R}_+$ . Prove that :
\[ \sqrt {x(y + 1)} + \sqrt {y(z + 1)} + \sqrt {z(x + 1)}\le \frac {3}{2}\sqrt {(x + 1)(y + 1)(z + 1)}\]
Search found 46 matches
- Tue Jun 01, 2021 11:42 am
- Forum: Algebra
- Topic: Inequality Marathon
- Replies: 9
- Views: 12121
- Tue Jun 01, 2021 11:32 am
- Forum: Algebra
- Topic: Inequality Marathon
- Replies: 9
- Views: 12121
Re: Problem 00
Let $a,b,c$ be positive real numbers satisfying $abc\geq1$ Prove that, \[\frac{a^2bc}{\sqrt{bc}+1}+\frac{b^2ca}{\sqrt{ca}+1}+\frac{c^2ab}{\sqrt{ab}+1}\geq\frac32\] Source : Adib Hasan's IMO Mock Collection 4 $\textbf{Solution 00}$ Let $a=x^2,b=y^2,c=z^2$. $\sum \frac{a^2bc}{\sqrt{bc}+1}=abc \sum \f...
- Fri May 14, 2021 11:21 am
- Forum: National Math Camp
- Topic: Problem - 03 - National Math Camp 2021 Mock Exam - "Functional equation, but not functioning well!"
- Replies: 3
- Views: 10814
Re: Problem - 03 - National Math Camp 2021 Mock Exam - "Functional equation, but not functioning well!"
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, \[f(f(f(x)+y))=f(x+y)+f(x)+y\] $P(x,f(y))\Rightarrow f(f(f(x)+f(y)))=f(x+f(y))+f(x)+f(y)$ $\Rightarrow f(x+f(y))=f(y+f(x))$ $\Rightarrow f(x+y)+f(y)+x=f(x+y)+f(x)+y$ $\Rightarrow f(x)=x+f(0)$ Plugging this into the...
- Mon May 10, 2021 11:26 pm
- Forum: College / University Level
- Topic: A question about derivatives
- Replies: 2
- Views: 9201
Re: A question about derivatives
Consider this following proof of a combinatorial identity i.e $n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ Proof: $(1+x)^n=\sum_{k=1}^{n} \binom{n}{k} x^k$ Differentiating both sides wrt x gives $n(1+x)^{n-1}=\sum_{k=1}^{n}k \binom{n}{k} x^{k-1}$ $x:=1 \Rightarrow n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ ...
- Sun May 09, 2021 5:00 pm
- Forum: National Math Camp
- Topic: National Math Camp Geometry Exam Problem-3
- Replies: 2
- Views: 8067
Re: National Math Camp Geometry Exam Problem-3
$Problem$ $3$. Let $ABC$ be a triangle with $BC$ being the longest side. Let $O$ be the circumcenter of $ABC$. $P$ is an arbitrary point on $BC$. The perpendicular bisector of $BP$ meet $AB$ at $Q$ and the perpendicular bisector of $PC$ meet $AC$ at $R$. Prove that $AQOR$ is cyclic. $\textbf{Soluti...
- Wed Apr 28, 2021 1:46 pm
- Forum: College / University Level
- Topic: Differentiation Marathon!
- Replies: 21
- Views: 26160
Re: Problem: 6
This is actually a self made problem, just for fun. Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ Note: There might be 2 derivatives. I don't know 2 derivatives are a thing or not. But just for the fun, we can do it, right? Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ $\Rightarrow y=\sqrt{x+y} ...
- Fri Apr 23, 2021 11:26 pm
- Forum: College / University Level
- Topic: Differentiation Marathon!
- Replies: 21
- Views: 26160
Re: Differentiation Marathon!
I have a different approach, let, $\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}} = y$ Then $\sqrt{x+y}=y$ Or,$x+y=y^2$ Or,$y^2-y-x=0$ Or,$y=\frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-x)}}{(2)(1)}$ Or,$y=\frac{1\pm \sqrt{1+4x}}{2}$ Differentiating on both sides we get, $\frac{d}{dx}y=\pm \frac{1}{\sqrt{1+4x}} $ Which i...
- Fri Apr 23, 2021 1:50 am
- Forum: College / University Level
- Topic: Differentiation Marathon!
- Replies: 21
- Views: 26160
Re: Problem: 6
This is actually a self made problem, just for fun. Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ Note: There might be 2 derivatives. I don't know 2 derivatives are a thing or not. But just for the fun, we can do it, right? Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ $\Rightarrow y=\sqrt{x+y} ...
- Thu Apr 22, 2021 2:07 pm
- Forum: Number Theory
- Topic: Need to confirm if this is right
- Replies: 2
- Views: 6903
Re: Need to confirm if this is right
Problem: Find all positive integers $x,y$ such that $p^x-y^p=1$ where $p$ is prime.(Czech Slovakia 1996) Solution:(Need to confirm :oops:) We can rewrite this as $p^x=1+y^p$ Now $p|1+y$ otherwise $1+y=1$ or $y=0$ which is not positive integer So, Applying LTE we can get $\nu_{p}(1+y)=x-1$. Since th...
- Wed Apr 21, 2021 3:26 pm
- Forum: National Math Camp
- Topic: National Camp Exam 2018 P3
- Replies: 4
- Views: 5872
Re: National Camp Exam 2018 P3
No, Telescoping