a) $a,b,c,d$ are positive reals such that $abcd=1$. Prove that \[\sum_{cyc} \frac{1+ab}{1+a}\geq 4.\](b)In a scalene triangle $ABC$, $\angle BAC =120^{\circ}$. The bisectors of angles $A,B,C$ meets the opposite sides in $P,Q,R$ respectively. Prove that the circle on $QR$ as diameter passes through the point $P$.
Problem
Source: NMTC 2017 Junior P4
Tags: inequalities, geometry
Mr.Techworm
28.10.2017 17:16
I don't see where the $$abcd=1$$comes in to picture. I did it through simple rerangement without using $$abcd=1$$.
RC.
28.10.2017 18:07
TheDarkPrince wrote: (b)In a scalene triangle $ABC$, $\angle BAC =120^{\circ}$. The bisectors of angles $A,B,C$ meets the opposite sides in $P,Q,R$ respectively. Prove that the circle on $QR$ as diameter passes through the point $P$.
Let \(I\) be the incenter of \(\Delta ABC\) and \(R' \in AB\) be a point such that \(PR'\) is the bisector of \(\angle APB\), \(PR' \cap BI \equiv M\). We have: \(\angle BIC = 150^{\circ} \Rightarrow \angle BIR = 30^{\circ} - (1)\) Also, \(M\) is the incenter of \(\Delta APB \Rightarrow \angle IMR' = 120^{\circ} \Rightarrow AIMR'\) is cyclic $\Rightarrow \angle MIR' = \angle MAB = 30^{\circ} = \angle BIR = \angle MIR$ $\Rightarrow R \equiv R'$ and thus \(PR\) bisects \(\angle APB\) similarly, \(PQ\) bisects \(\angle APC \Rightarrow \angle QPR = 90^{\circ}\)
Is the world running out of problems or is it just they believe in revision?
bel.jad5
29.10.2017 00:16
Mr.Techworm wrote: I don't see where the $$abcd=1$$comes in to picture. I did it through simple rerangement without using $$abcd=1$$. The inequality is wrong for (1,0,0,0). So something is wrong with your proof
mira74
29.10.2017 05:39
bel.jad5 wrote: Mr.Techworm wrote: I don't see where the $$abcd=1$$comes in to picture. I did it through simple rerangement without using $$abcd=1$$. The inequality is wrong for (1,0,0,0). So something is wrong with your proof TheDarkPrince wrote: a) $a,b,c,d$ are positive reals
bel.jad5
29.10.2017 09:40
Hmm...try (1,1/10000000,1/0000000,1/10000000) then
Grotex
29.10.2017 09:44
bel.jad5 wrote: Hmm...try (1,1/10000000,1/0000000,1/10000000) then $abcd=1,$ but $1*1/10000000*1/0000000*1/10000000\neq 1$
bel.jad5
30.10.2017 10:26
Grotex wrote: bel.jad5 wrote: Hmm...try (1,1/10000000,1/0000000,1/10000000) then $abcd=1,$ but $1*1/10000000*1/0000000*1/10000000\neq 1$ You didnt follow my train of thought Mr.Techworm wrote: I don't see where the $$abcd=1$$comes in to picture. I did it through simple rerangement without using $$abcd=1$$.
RagvaloD
30.10.2017 11:05
$\sum_{cyc} \frac{1+ab}{1+a}=\frac{1+ab}{1+a}+\frac{1+cd}{1+c}+\frac{1+bc}{1+b}+\frac{1+ad}{1+d}=(1+ab)( \frac{1}{1+a}+\frac{1}{ab(1+c)})+(1+bc) (\frac{1}{1+b}+\frac{1}{bc(1+d)}) \geq 4\frac{1+ab}{1+a+ab+abc}+4\frac{1+bc}{1+b+bc+bcd}=4\frac{1+ab}{1+a+ab+abc}+4\frac{a(1+bc)}{1+a+ab+abc}=4$
Mr.Techworm
30.10.2017 18:20
RagvaloD wrote: $\sum_{cyc} \frac{1+ab}{1+a}=\frac{1+ab}{1+a}+\frac{1+cd}{1+c}+\frac{1+bc}{1+b}+\frac{1+ad}{1+d}=(1+ab)( \frac{1}{1+a}+\frac{1}{ab(1+c)})+(1+bc) (\frac{1}{1+b}+\frac{1}{bc(1+d)}) \geq 4\frac{1+ab}{1+a+ab+abc}+4\frac{1+bc}{1+b+bc+bcd}=4\frac{1+ab}{1+a+ab+abc}+4\frac{a(1+bc)}{1+a+ab+abc}=4$ Beautiful RagvaloD
MathAllTheWay
30.10.2017 19:00
a.) Hint: Use the substitution $ a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{w}, d= \frac{w}{x} $ and then apply the AM-GM inequality
sqing
31.10.2017 06:38
RagvaloD wrote: $\sum_{cyc} \frac{1+ab}{1+a}=\frac{1+ab}{1+a}+\frac{1+cd}{1+c}+\frac{1+bc}{1+b}+\frac{1+ad}{1+d}=(1+ab)( \frac{1}{1+a}+\frac{1}{ab(1+c)})+(1+bc) (\frac{1}{1+b}+\frac{1}{bc(1+d)}) \geq 4\frac{1+ab}{1+a+ab+abc}+4\frac{1+bc}{1+b+bc+bcd}=4\frac{1+ab}{1+a+ab+abc}+4\frac{a(1+bc)}{1+a+ab+abc}=4$ $\frac{1+ab}{1+a}+\frac{1+cd}{1+c}+\frac{1+bc}{1+b}+\frac{1+ad}{1+d}$ $=(1+ab)\left( \frac{1}{1+a}+\frac{1}{ab(1+c)}\right)+(1+bc)\left (\frac{1}{1+b}+\frac{1}{bc(1+d)}\right)$ $ \geq \frac{4(1+ab)}{1+a+ab+abc}+\frac{4(1+bc)}{1+b+bc+bcd}$ $=\frac{4(1+ab)}{1+a+ab+abc}+\frac{4a(1+bc)}{1+a+ab+abc}=4$ Very nice.
sqing
02.11.2017 04:26
Let $a,b,c,d$ be positive real numbers such that $abcd\geq1$. Prove that$$\frac{a+b}{a+1}+\frac{b+c}{b+1}+\frac{c+d}{c+1}+\frac{d+a}{d+1}\leq a + b+c + d$$Mathematical Reflections 4 (2017)