Prove $frac{a+b}{sqrt{c}}+frac{a+c}{sqrt{b}}+ frac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c})$
Could anyone advise me on how to prove this inequality:
$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$
where $a,b,c $ are any positive real numbers.
Do I use the AM-GM inequality somewhere?
Thank you.
algebra-precalculus inequality
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
add a comment |
Could anyone advise me on how to prove this inequality:
$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$
where $a,b,c $ are any positive real numbers.
Do I use the AM-GM inequality somewhere?
Thank you.
algebra-precalculus inequality
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01
add a comment |
Could anyone advise me on how to prove this inequality:
$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$
where $a,b,c $ are any positive real numbers.
Do I use the AM-GM inequality somewhere?
Thank you.
algebra-precalculus inequality
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
Could anyone advise me on how to prove this inequality:
$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$
where $a,b,c $ are any positive real numbers.
Do I use the AM-GM inequality somewhere?
Thank you.
algebra-precalculus inequality
algebra-precalculus inequality
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
edited Dec 28 '18 at 16:44
Martin Sleziak
44.6k8115271
44.6k8115271
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
asked Nov 4 '15 at 0:54
Alexy VincenzoAlexy Vincenzo
2,1753925
2,1753925
Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01
add a comment |
Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01
Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01
Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01
add a comment |
1 Answer
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From AM-GM you have
$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.
edited Nov 4 '15 at 1:28
Kaster
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
From AM-GM you have
$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.
edited Nov 4 '15 at 1:28
Kaster
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
add a comment |
From AM-GM you have
$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.
edited Nov 4 '15 at 1:28
Kaster
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
add a comment |
From AM-GM you have
$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.
edited Nov 4 '15 at 1:28
Kaster
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
From AM-GM you have
$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.
edited Nov 4 '15 at 1:28
Kaster
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
edited Nov 4 '15 at 1:28
Kaster
9,04221729
edited Nov 4 '15 at 1:28
Kaster
9,04221729
edited Nov 4 '15 at 1:28
Kaster
9,04221729
9,04221729
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
answered Nov 4 '15 at 1:25
clarkclark
12.9k22657
12.9k22657
add a comment |
add a comment |
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Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01