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

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The discussion centers on proving that 1/a + 1/b + 1/c + 1/d > 1 under specific conditions for positive variables a, b, c, and d. Participants debate the validity of different proof methods, including Jensen's Inequality, and express concerns about the correctness of initial solutions. One contributor suggests that simpler approaches may help students grasp the concepts better, while another emphasizes the need for rigorous mathematical proof rather than intuitive reasoning. The conversation highlights the complexity of the problem and the importance of considering all variables in the proof. Ultimately, the need for a definitive solution using Jensen's Inequality remains a focal point.
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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 $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>1$.
 
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anemone said:
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 $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>1$.
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----

 
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other solution :
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


 
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Albert said:
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! :)

Albert said:
other solution :
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...:confused: 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.
 
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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
 
Albert said:
other solution :
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

a = 1.2, b= 4.7 is a contradiction
 
kaliprasad said:
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
 
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Albert said:
$\dfrac {1}{1.2}+\dfrac {1}{4.7}>1$ 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.
 
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
 
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  • #10
Solution of other using Jensen's Inequality:

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.
 
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