Can Jensen's Inequality Prove 1/a+1/b+1/c+1/d>1?

[anemone]

Let $a,\,b,\,c,\,d>0$ such that $a<2$, $a+b<6$, $a+b+c<12$ and $a+b+c+d<24$.

Prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>1$.

 

[Albert1]

Let $a,\,b,\,c,\,d>0$ such that $a<2---(1)$, $a+b<6---(2)$, $a+b+c<12---(3)$ and $a+b+c+d<24---(4)$.

Prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>1$.

Spoiler

from (1):$\dfrac {1}{a}>\dfrac {1}{2}---(5)$
using $AP\geq GP$
from (2)$6>a+b\geq 2\sqrt{ab}$
$\therefore ab< 9, or,\,\, \dfrac {1}{ab}>\dfrac {1}{9}$
$\dfrac {1}{a}+\dfrac {1}{b}\geq 2\sqrt {\dfrac {1}{ab}}>\dfrac {2}{3}--(6)$
from (3)$12>a+b+c\geq 3\sqrt[3]{abc}$
$\therefore abc< 64, or,\,\, \dfrac {1}{abc}>\dfrac {1}{64}$
$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}\geq 3\sqrt [3]{\dfrac{1}{abc}}>\dfrac {3}{4}---(7)$
to be continued----

 

[Albert1]

other solution :

Spoiler

range of $a:1<a<2$
range of $b:1<b<4$
range of $c: 1<c<6$
range of $d : 1<d<12$
so $\dfrac {1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac {1}{d}>\dfrac {6+3+2+1}{12}=1$
for if anyone of $a,b,c,d\leq 1$ then there is no need to prove

 

[anemone]

Spoiler

from (1):$\dfrac {1}{a}>\dfrac {1}{2}---(5)$
using $AP\geq GP$
from (2)$6>a+b\geq 2\sqrt{ab}$
$\therefore ab< 9, or,\,\, \dfrac {1}{ab}>\dfrac {1}{9}$
$\dfrac {1}{a}+\dfrac {1}{b}\geq 2\sqrt {\dfrac {1}{ab}}>\dfrac {2}{3}--(6)$
from (3)$12>a+b+c\geq 3\sqrt[3]{abc}$
$\therefore abc< 64, or,\,\, \dfrac {1}{abc}>\dfrac {1}{64}$
$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}\geq 3\sqrt [3]{\dfrac{1}{abc}}>\dfrac {3}{4}---(7)$
to be continued----

Hi Albert,

I hope you would remember to finish the proof, but I will wait! :)

other solution :

Spoiler

range of $a:1<a<2$
range of $b:1<b<4$
range of $c: 1<c<6$
range of $d : 1<d<12$
so $\dfrac {1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac {1}{d}>\dfrac {6+3+2+1}{12}=1$
for if anyone of $a,b,c,d\leq 1$ then there is no need to prove

Hmm...I cannot say this proof is not correct, but I have the feeling that something is amiss... chances are I could be wrong though.

I will post back for the solution using the Jensen's Inequality method, not today but soon, I promise.

 

[Albert1]

I am sure my second solution is correct ,and nothing is amiss
I usually use simpler way to solve a math problem , this will give my students a better understanding
in fact there is something amiss in my first solution , for using $AP\geq GP$,
the ranges $0<a,b,c,d \leq 1$ have also taken into consideration
within this range ,the statement is always true there is no need to prove

 

[kaliprasad]

other solution :

Spoiler

range of $a:1<a<2$
range of $b:1<b<4$
range of $c: 1<c<6$
range of $d : 1<d<12$
so $\dfrac {1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac {1}{d}>\dfrac {6+3+2+1}{12}=1$
for if anyone of $a,b,c,d\leq 1$ then there is no need to prove

Spoiler

a = 1.2, b= 4.7 is a contradiction

 

[Albert1]

Spoiler

a = 1.2, b= 4.7 is a contradiction

$\dfrac {1}{1.2}+\dfrac {1}{4.7}>1$ and there is no need to take c,d into consideration
when anyone of a,b,c,d is bigger than 1 and very close to 1 then its reciprocal very close to 1
for example if you take b=4 then $\dfrac {1}{b}=0.25$ ,and $a$ must bigger then $\dfrac {4}{3}>1.2 $ for $\dfrac {3}{4}=0.75$ ,the sum of $\dfrac {1}{a}+\dfrac {1}{b}=1$
may be you can help me to find better ranges for a,b,c,d

 

[kaliprasad]

$\dfrac {1}{1.2}+\dfrac {1}{4.7}>1$ and there is no need to take c,d into consideration
when anyone of a,b,c,d is bigger than 1 and very close to 1 then its reciprocal very close to 1
for example if you take b=4 then $\dfrac {1}{b}=0.25$ ,and $a$ must bigger then $\dfrac {4}{3}>1.2 $ for $\dfrac {3}{4}=0.75$ ,the sum of $\dfrac {1}{a}+\dfrac {1}{b}=1$
may be you can help me to find better ranges for a,b,c,d

what I meant that I have provided a case which contradicts your claim if b < 4 and your solution is inituitive and not correct mathemetically This is no way to provide a solution based on gut feeling.

 

[Albert1]

if a,b,c,d all should be taken into consideration ,then a,b,c,d must be as bigger as possible ,
that is 1/a, 1/b, 1/c and 1/d as smaller as possible (and "a" affects most)
for example even if we let b=5 ,then "a" very close to 5/4 =1.25>1.2
I admit my solution is not strict ,but sometimes by observation we can make the original problem easier for studens to understand

 

[anemone]

Solution of other using Jensen's Inequality:

Spoiler

Applying Jensen's Inequality to the function $f(x)=\dfrac{1}{x}$ for $x>0$ gives

$\begin{align*}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}&=\dfrac{1}{2}f\left(\dfrac{a}{2}\right)+\dfrac{1}{4}f\left(\dfrac{b}{4}\right)+\dfrac{1}{6}f\left(\dfrac{c}{6}\right)+\dfrac{1}{12}f\left(\dfrac{d}{12}\right)\\&>f\left(\dfrac{a}{4}+\dfrac{b}{16}+\dfrac{c}{36}+\dfrac{d}{144}\right)\\&=\dfrac{144}{36a+9b+4c+d}\\&=\dfrac{144}{27a+5(a+b)+3(a+b+c)+(a+b+c+d)}\\&>\dfrac{144}{27(2)+5(6)+3(12)+24}=1\end{align*}$

and we're hence done.