Question

inequality
part b

Answer 1

@asnaseer @Zarkon @FoolForMath @Callisto any takers on a tough one?

Answer 2

the solution is so long !!!!
just multiple a+b+c to both side of equation and replace ab+ac with 1/3-bc and do this for other fraction .

Answer 3

and then?

Answer 4

From roya's idea, I got the solution:
Begin with \(\frac{a}{a^2-bc+1}\). Multiply this by (a+b+c)/(a+b+c):
\[\frac{a}{a^2-bc+1}\cdot \frac{a+b+c}{a+b+c}=\frac{a^2+ab+ac}{a^2-bc+1}\cdot \frac{1}{a+b+c}\\
=\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}\cdot \frac{1}{a+b+c}\\
\geq\frac{1}{3}\cdot\frac{1}{a+b+c}\]
I think you can verify that for any positive quantity x, \(\frac{x+a}{x+b}\geq\frac{a}{b} \text{for }a<b\).
All that's left is to verify that \(a^2-bc\geq0\) for a, b, c>0.
But I can't see how to do this. :(

Answer 5

if u solve all fraction there are 3 part like a^2-bc and 2 other like this . i think with 3 part u can solve it but i didn't have time for doing this

Answer 6

if a=1/10, b=2, c=4/63 such that ab+bc+ca=1/3, but \[a^2-bc <0\]

Answer 7

if u see all 3 fraction with each other u get the answer .

Answer 8

if a^2 -bc>=0 , b^2-ac>=0 and c^2-ab>=0,
I get the answer.
but a^2 -bc may <0

Answer 9

a^2+b^2+c^2>ab+bc+cb

Answer 10

we can use (x+a)/(x+b)>=a/b if x<0 ?

Answer 11

it's not depend on x :
look at this

Answer 12

\[(x+a)/(x+b)>a/b \rightarrow bx+ab>ax+ab \rightarrow bx>ax \rightarrow b>a \]

Answer 13

now 1/3 < 1 , but a^2 -bc < 0.
It cannot say (a^2 -bc+1/3)/(a^2 -bc+1) >= 1/3

Answer 14

bx > ax
b<a if x<0

Answer 15

ya you are correct .

Answer 16

i think u have to use the answer of part a

Answer 17

again we have to prove a^2>bc ? how ???

Answer 18

i get it !!!
look at this :

Answer 19

\[(a^2+ab+ac)/((a^2+ab+ac)+2/3) >0\]

Answer 20

(a^2+ab+ac)/((a^2+ab+ac)+2/3)>0 but not >1/3

Answer 21

i find the solution :)

Answer 22

how u reach the 3rd step ?

Answer 23

i get it
fantastic

Answer 24

Was the 3rd step Cauchy-Schwarz? I don't recall Cauchy-Schwarz looking like that...

Answer 25

\[\left( \sum_{i=1}^{n}x_{i}^2 \right)\left( \sum_{i=1}^{n}y_{i}^2 \right) \ge\left( \sum_{i=1}^{n}x_{i}y_i \right)^2\]
rewrite \[x_i \to \sqrt{x_i}\] and \[y_i \to c_i / \sqrt{x_i}\]
we have
\[\left( \sum_{i=1}^{n}x_{i} \right)\left( \sum_{i=1}^{n}c_{i}^2/x_i \right) \ge\left( \sum_{i=1}^{n}c_i \right)^2\]
\[\left( \sum_{i=1}^{n}c_{i}^2/x_i \right) \ge\left( \sum_{i=1}^{n}c_i \right)^2 / \left( \sum_{i=1}^{n}x_{i} \right)\]