How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) ge 3/2$ [duplicate] The 2019 Stack Overflow Developer Survey Results Are InProof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$Proof of Nesbitt's Inequality: $fracab+c+fracbc+a+fracca+bge frac32$?how to prove this inequality?how to solve a inequality?Prove this inequality with $xyzle 1$How to show the inequality is strict?An inequality about the sum of distances between points : same color $le$ different colors?How to prove this inequalityMaybe this inequality holds? $x!-y!>x^n$?Prove this by inequality with four variables inequalityA tricky integral inequalityClueless as to how to solve this gamma function differential inequality
How to change the limits of integration
I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?
What is the meaning of Triage in Cybersec world?
Is there a name of the flying bionic bird?
Is bread bad for ducks?
Understanding the implication of what "well-defined" means for the operation in quotient group
What is the purpose of the constant in the probability density function
Inline version of a function returns different value than non-inline version
Does a dangling wire really electrocute me if I'm standing in water?
Why is my p-value correlated to difference between means in two sample tests?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
How to deal with fear of taking dependencies
How to create dashed lines/arrows in Illustrator
A poker game description that does not feel gimmicky
How to manage monthly salary
JSON.serialize: is it possible to suppress null values of a map?
How to answer pointed "are you quitting" questioning when I don't want them to suspect
Deadlock Graph and Interpretation, solution to avoid
The difference between dialogue marks
Pristine Bit Checking
Access elements in std::string where positon of string is greater than its size
Is flight data recorder erased after every flight?
Why is Grand Jury testimony secret?
How was Skylab's orbit inclination chosen?
How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) ge 3/2$ [duplicate]
The 2019 Stack Overflow Developer Survey Results Are InProof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$Proof of Nesbitt's Inequality: $fracab+c+fracbc+a+fracca+bge frac32$?how to prove this inequality?how to solve a inequality?Prove this inequality with $xyzle 1$How to show the inequality is strict?An inequality about the sum of distances between points : same color $le$ different colors?How to prove this inequalityMaybe this inequality holds? $x!-y!>x^n$?Prove this by inequality with four variables inequalityA tricky integral inequalityClueless as to how to solve this gamma function differential inequality
$begingroup$
This question already has an answer here:
Proof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$
5 answers
Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c ge frac32cdot [(a+b)(a+c)(b+c)]^frac13$$
Hence or otherwise prove:
$$colorbluefracab+c+fracba+c+fracca+bge frac32$$
inequality
edited Mar 30 at 16:54
Dr. Mathva
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
$endgroup$
marked as duplicate by Lord Shark the Unknown, Javi, darij grinberg, Lee David Chung Lin, Tianlalu Apr 3 at 1:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$
5 answers
Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c ge frac32cdot [(a+b)(a+c)(b+c)]^frac13$$
Hence or otherwise prove:
$$colorbluefracab+c+fracba+c+fracca+bge frac32$$
inequality
edited Mar 30 at 16:54
Dr. Mathva
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
$endgroup$
marked as duplicate by Lord Shark the Unknown, Javi, darij grinberg, Lee David Chung Lin, Tianlalu Apr 3 at 1:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56
add a comment |
$begingroup$
This question already has an answer here:
Proof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$
5 answers
Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c ge frac32cdot [(a+b)(a+c)(b+c)]^frac13$$
Hence or otherwise prove:
$$colorbluefracab+c+fracba+c+fracca+bge frac32$$
inequality
edited Mar 30 at 16:54
Dr. Mathva
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
$endgroup$
This question already has an answer here:
Proof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$
5 answers
Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c ge frac32cdot [(a+b)(a+c)(b+c)]^frac13$$
Hence or otherwise prove:
$$colorbluefracab+c+fracba+c+fracca+bge frac32$$
This question already has an answer here:
Proof of the inequality $fracab+c+fracba+c+fracca+b geq frac32$
5 answers
inequality
inequality
edited Mar 30 at 16:54
Dr. Mathva
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
edited Mar 30 at 16:54
Dr. Mathva
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
edited Mar 30 at 16:54
Dr. Mathva
3,493630
edited Mar 30 at 16:54
Dr. Mathva
3,493630
edited Mar 30 at 16:54
Dr. Mathva
3,493630
3,493630
asked Mar 30 at 11:47
M.YouM.You
21
asked Mar 30 at 11:47
M.YouM.You
21
asked Mar 30 at 11:47
M.YouM.You
21
21
marked as duplicate by Lord Shark the Unknown, Javi, darij grinberg, Lee David Chung Lin, Tianlalu Apr 3 at 1:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, Javi, darij grinberg, Lee David Chung Lin, Tianlalu Apr 3 at 1:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56
add a comment |
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Using AM-GM inequality:$$fracx_1+...+x_nngeq (x_1...x_n)^1/n$$
let $n=3$ and $x_1=a+b,x_2=b+c,x_3=c+a$:
$$frac2(a+b+c)3geq [(a+b)(b+c)(c+a)]^1/3$$
answered Mar 30 at 11:59
Xin FuXin Fu
34319
$endgroup$
add a comment |
$begingroup$
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=frac-x+y+z2$$
$$b=fracx-y+z2$$
$$c=fracx+y-z2$$
And we get
$$frac-x+y+z2x+fracx-y+z2y+fracx+y-z2zgeq frac32$$
Can you finish?
And we get $$fracxy+fracyx+fracyz+fraczy+fracxz+fraczxgeq 6$$
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
To minimize $fracab+c+fracbc+a+fracca+b$ , we need to have
$$
beginalign
0
&=deltaleft(fracab+c+fracbc+a+fracca+bright)\
&=left(frac1b+c-fracb(c+a)^2-fracc(a+b)^2right)delta a\
&+left(frac1c+a-fracc(a+b)^2-fraca(b+c)^2right)delta b\
&+left(frac1a+b-fraca(b+c)^2-fracb(c+a)^2right)delta ctag1
endalign
$$
for all $delta a,delta b,delta c$. That means
$$
beginalign
frac1a+b&=fraca(b+c)^2+fracb(c+a)^2tag2\
frac1b+c&=fracb(c+a)^2+fracc(a+b)^2tag3\
frac1c+a&=fracc(a+b)^2+fraca(b+c)^2tag4
endalign
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
frac1b+c+frac1a+b-frac1c+a=frac2b(c+a)^2tag5
$$
Add $frac2c+a$ and divide by $2(a+b+c)$:
$$
frac12(a+b+c)left(frac1b+c+frac1a+b+frac1c+aright)=frac1(c+a)^2tag6
$$
By symmetry,
$$
frac1(a+b)^2=frac1(b+c)^2=frac1(c+a)^2tag7
$$
from which we get $a=b=c$. Thus, we get
$$
fracab+c+fracbc+a+fracca+bgefrac32tag8
$$
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracab+c+fracba+c+fracca+bge frac32iff fraca+b+cb+c+fracb+a+ca+c+fracc+a+ba+bge frac32+3$$
$$iff big(a+b+cbig)cdot bigg(frac1b+c+frac1a+c+frac1a+bbigg)ge frac92iff colorbluefrac2cdot (a+b+c)3ge frac3frac1b+c+frac1a+c+frac1a+b$$ Which is trivial by the AM-HM inequality. Done!
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using AM-GM inequality:$$fracx_1+...+x_nngeq (x_1...x_n)^1/n$$
let $n=3$ and $x_1=a+b,x_2=b+c,x_3=c+a$:
$$frac2(a+b+c)3geq [(a+b)(b+c)(c+a)]^1/3$$
answered Mar 30 at 11:59
Xin FuXin Fu
34319
$endgroup$
add a comment |
$begingroup$
Using AM-GM inequality:$$fracx_1+...+x_nngeq (x_1...x_n)^1/n$$
let $n=3$ and $x_1=a+b,x_2=b+c,x_3=c+a$:
$$frac2(a+b+c)3geq [(a+b)(b+c)(c+a)]^1/3$$
answered Mar 30 at 11:59
Xin FuXin Fu
34319
$endgroup$
add a comment |
$begingroup$
Using AM-GM inequality:$$fracx_1+...+x_nngeq (x_1...x_n)^1/n$$
let $n=3$ and $x_1=a+b,x_2=b+c,x_3=c+a$:
$$frac2(a+b+c)3geq [(a+b)(b+c)(c+a)]^1/3$$
answered Mar 30 at 11:59
Xin FuXin Fu
34319
$endgroup$
Using AM-GM inequality:$$fracx_1+...+x_nngeq (x_1...x_n)^1/n$$
let $n=3$ and $x_1=a+b,x_2=b+c,x_3=c+a$:
$$frac2(a+b+c)3geq [(a+b)(b+c)(c+a)]^1/3$$
answered Mar 30 at 11:59
Xin FuXin Fu
34319
answered Mar 30 at 11:59
Xin FuXin Fu
34319
answered Mar 30 at 11:59
Xin FuXin Fu
34319
answered Mar 30 at 11:59
Xin FuXin Fu
34319
34319
add a comment |
add a comment |
$begingroup$
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=frac-x+y+z2$$
$$b=fracx-y+z2$$
$$c=fracx+y-z2$$
And we get
$$frac-x+y+z2x+fracx-y+z2y+fracx+y-z2zgeq frac32$$
Can you finish?
And we get $$fracxy+fracyx+fracyz+fraczy+fracxz+fraczxgeq 6$$
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=frac-x+y+z2$$
$$b=fracx-y+z2$$
$$c=fracx+y-z2$$
And we get
$$frac-x+y+z2x+fracx-y+z2y+fracx+y-z2zgeq frac32$$
Can you finish?
And we get $$fracxy+fracyx+fracyz+fraczy+fracxz+fraczxgeq 6$$
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=frac-x+y+z2$$
$$b=fracx-y+z2$$
$$c=fracx+y-z2$$
And we get
$$frac-x+y+z2x+fracx-y+z2y+fracx+y-z2zgeq frac32$$
Can you finish?
And we get $$fracxy+fracyx+fracyz+fraczy+fracxz+fraczxgeq 6$$
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=frac-x+y+z2$$
$$b=fracx-y+z2$$
$$c=fracx+y-z2$$
And we get
$$frac-x+y+z2x+fracx-y+z2y+fracx+y-z2zgeq frac32$$
Can you finish?
And we get $$fracxy+fracyx+fracyz+fraczy+fracxz+fraczxgeq 6$$
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
edited Mar 30 at 12:02
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Mar 30 at 11:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
78.8k42867
add a comment |
add a comment |
$begingroup$
To minimize $fracab+c+fracbc+a+fracca+b$ , we need to have
$$
beginalign
0
&=deltaleft(fracab+c+fracbc+a+fracca+bright)\
&=left(frac1b+c-fracb(c+a)^2-fracc(a+b)^2right)delta a\
&+left(frac1c+a-fracc(a+b)^2-fraca(b+c)^2right)delta b\
&+left(frac1a+b-fraca(b+c)^2-fracb(c+a)^2right)delta ctag1
endalign
$$
for all $delta a,delta b,delta c$. That means
$$
beginalign
frac1a+b&=fraca(b+c)^2+fracb(c+a)^2tag2\
frac1b+c&=fracb(c+a)^2+fracc(a+b)^2tag3\
frac1c+a&=fracc(a+b)^2+fraca(b+c)^2tag4
endalign
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
frac1b+c+frac1a+b-frac1c+a=frac2b(c+a)^2tag5
$$
Add $frac2c+a$ and divide by $2(a+b+c)$:
$$
frac12(a+b+c)left(frac1b+c+frac1a+b+frac1c+aright)=frac1(c+a)^2tag6
$$
By symmetry,
$$
frac1(a+b)^2=frac1(b+c)^2=frac1(c+a)^2tag7
$$
from which we get $a=b=c$. Thus, we get
$$
fracab+c+fracbc+a+fracca+bgefrac32tag8
$$
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
$begingroup$
To minimize $fracab+c+fracbc+a+fracca+b$ , we need to have
$$
beginalign
0
&=deltaleft(fracab+c+fracbc+a+fracca+bright)\
&=left(frac1b+c-fracb(c+a)^2-fracc(a+b)^2right)delta a\
&+left(frac1c+a-fracc(a+b)^2-fraca(b+c)^2right)delta b\
&+left(frac1a+b-fraca(b+c)^2-fracb(c+a)^2right)delta ctag1
endalign
$$
for all $delta a,delta b,delta c$. That means
$$
beginalign
frac1a+b&=fraca(b+c)^2+fracb(c+a)^2tag2\
frac1b+c&=fracb(c+a)^2+fracc(a+b)^2tag3\
frac1c+a&=fracc(a+b)^2+fraca(b+c)^2tag4
endalign
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
frac1b+c+frac1a+b-frac1c+a=frac2b(c+a)^2tag5
$$
Add $frac2c+a$ and divide by $2(a+b+c)$:
$$
frac12(a+b+c)left(frac1b+c+frac1a+b+frac1c+aright)=frac1(c+a)^2tag6
$$
By symmetry,
$$
frac1(a+b)^2=frac1(b+c)^2=frac1(c+a)^2tag7
$$
from which we get $a=b=c$. Thus, we get
$$
fracab+c+fracbc+a+fracca+bgefrac32tag8
$$
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
$endgroup$
add a comment |
$begingroup$
To minimize $fracab+c+fracbc+a+fracca+b$ , we need to have
$$
beginalign
0
&=deltaleft(fracab+c+fracbc+a+fracca+bright)\
&=left(frac1b+c-fracb(c+a)^2-fracc(a+b)^2right)delta a\
&+left(frac1c+a-fracc(a+b)^2-fraca(b+c)^2right)delta b\
&+left(frac1a+b-fraca(b+c)^2-fracb(c+a)^2right)delta ctag1
endalign
$$
for all $delta a,delta b,delta c$. That means
$$
beginalign
frac1a+b&=fraca(b+c)^2+fracb(c+a)^2tag2\
frac1b+c&=fracb(c+a)^2+fracc(a+b)^2tag3\
frac1c+a&=fracc(a+b)^2+fraca(b+c)^2tag4
endalign
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
frac1b+c+frac1a+b-frac1c+a=frac2b(c+a)^2tag5
$$
Add $frac2c+a$ and divide by $2(a+b+c)$:
$$
frac12(a+b+c)left(frac1b+c+frac1a+b+frac1c+aright)=frac1(c+a)^2tag6
$$
By symmetry,
$$
frac1(a+b)^2=frac1(b+c)^2=frac1(c+a)^2tag7
$$
from which we get $a=b=c$. Thus, we get
$$
fracab+c+fracbc+a+fracca+bgefrac32tag8
$$
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
$endgroup$
To minimize $fracab+c+fracbc+a+fracca+b$ , we need to have
$$
beginalign
0
&=deltaleft(fracab+c+fracbc+a+fracca+bright)\
&=left(frac1b+c-fracb(c+a)^2-fracc(a+b)^2right)delta a\
&+left(frac1c+a-fracc(a+b)^2-fraca(b+c)^2right)delta b\
&+left(frac1a+b-fraca(b+c)^2-fracb(c+a)^2right)delta ctag1
endalign
$$
for all $delta a,delta b,delta c$. That means
$$
beginalign
frac1a+b&=fraca(b+c)^2+fracb(c+a)^2tag2\
frac1b+c&=fracb(c+a)^2+fracc(a+b)^2tag3\
frac1c+a&=fracc(a+b)^2+fraca(b+c)^2tag4
endalign
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
frac1b+c+frac1a+b-frac1c+a=frac2b(c+a)^2tag5
$$
Add $frac2c+a$ and divide by $2(a+b+c)$:
$$
frac12(a+b+c)left(frac1b+c+frac1a+b+frac1c+aright)=frac1(c+a)^2tag6
$$
By symmetry,
$$
frac1(a+b)^2=frac1(b+c)^2=frac1(c+a)^2tag7
$$
from which we get $a=b=c$. Thus, we get
$$
fracab+c+fracbc+a+fracca+bgefrac32tag8
$$
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
answered Mar 30 at 17:45
robjohn♦robjohn
270k27313642
270k27313642
add a comment |
add a comment |
$begingroup$
Hint:
$$fracab+c+fracba+c+fracca+bge frac32iff fraca+b+cb+c+fracb+a+ca+c+fracc+a+ba+bge frac32+3$$
$$iff big(a+b+cbig)cdot bigg(frac1b+c+frac1a+c+frac1a+bbigg)ge frac92iff colorbluefrac2cdot (a+b+c)3ge frac3frac1b+c+frac1a+c+frac1a+b$$ Which is trivial by the AM-HM inequality. Done!
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracab+c+fracba+c+fracca+bge frac32iff fraca+b+cb+c+fracb+a+ca+c+fracc+a+ba+bge frac32+3$$
$$iff big(a+b+cbig)cdot bigg(frac1b+c+frac1a+c+frac1a+bbigg)ge frac92iff colorbluefrac2cdot (a+b+c)3ge frac3frac1b+c+frac1a+c+frac1a+b$$ Which is trivial by the AM-HM inequality. Done!
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracab+c+fracba+c+fracca+bge frac32iff fraca+b+cb+c+fracb+a+ca+c+fracc+a+ba+bge frac32+3$$
$$iff big(a+b+cbig)cdot bigg(frac1b+c+frac1a+c+frac1a+bbigg)ge frac92iff colorbluefrac2cdot (a+b+c)3ge frac3frac1b+c+frac1a+c+frac1a+b$$ Which is trivial by the AM-HM inequality. Done!
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
$endgroup$
Hint:
$$fracab+c+fracba+c+fracca+bge frac32iff fraca+b+cb+c+fracb+a+ca+c+fracc+a+ba+bge frac32+3$$
$$iff big(a+b+cbig)cdot bigg(frac1b+c+frac1a+c+frac1a+bbigg)ge frac92iff colorbluefrac2cdot (a+b+c)3ge frac3frac1b+c+frac1a+c+frac1a+b$$ Which is trivial by the AM-HM inequality. Done!
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
answered Mar 30 at 17:13
Dr. MathvaDr. Mathva
3,493630
3,493630
add a comment |
add a comment |
$begingroup$
The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:50
$begingroup$
Also, the first inequality you are asked to prove follows from the AM-GM inequality ($fracx+y+z3ge sqrt[3]xyz$ for $x,y,zge 0$).
$endgroup$
– Minus One-Twelfth
Mar 30 at 11:56