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Let $a>b$ and $ab=1$ show that $frac{a^2+b^2}{a-b} geq 2sqrt{2}$
up vote
1
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Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$
My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
But this isn't very helpful.
inequality proof-writing
edited 11 hours ago
greedoid
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
add a comment |
up vote
1
down vote
favorite
Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$
My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
But this isn't very helpful.
inequality proof-writing
edited 11 hours ago
greedoid
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$
My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
But this isn't very helpful.
inequality proof-writing
edited 11 hours ago
greedoid
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
Let $a>b$ and $ab=1$. Show that $$frac{a^2+b^2}{a-b} geq 2sqrt{2}$$
My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
By AG inequality we know that $$frac{a^2+b^2}{2}geq sqrt{a^2b^2}=ab=1$$
But this isn't very helpful.
inequality proof-writing
inequality proof-writing
edited 11 hours ago
greedoid
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
edited 11 hours ago
greedoid
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
edited 11 hours ago
greedoid
34.3k114488
edited 11 hours ago
greedoid
34.3k114488
edited 11 hours ago
greedoid
34.3k114488
34.3k114488
asked 12 hours ago
Lovro Sindičić
244216
asked 12 hours ago
Lovro Sindičić
244216
asked 12 hours ago
Lovro Sindičić
244216
244216
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
You may proceed as follows using AM-GM:
$$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$
answered 11 hours ago
trancelocation
8,0561519
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up vote
1
down vote
Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.
answered 11 hours ago
José Carlos Santos
139k18111203
add a comment |
up vote
0
down vote
Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.
answered 11 hours ago
TheSimpliFire
11.6k62256
add a comment |
up vote
0
down vote
Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$
which is the same as $$(x-sqrt{2})^2geq 0$$
and we are done.
answered 11 hours ago
greedoid
34.3k114488
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You may proceed as follows using AM-GM:
$$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$
answered 11 hours ago
trancelocation
8,0561519
add a comment |
up vote
4
down vote
accepted
You may proceed as follows using AM-GM:
$$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$
answered 11 hours ago
trancelocation
8,0561519
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You may proceed as follows using AM-GM:
$$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$
answered 11 hours ago
trancelocation
8,0561519
You may proceed as follows using AM-GM:
$$frac{a^2+b^2}{a-b} = frac{a^2+b^2 - 2ab + 2}{a-b} = frac{(a-b)^2 + 2}{a-b}= (a-b)+ frac{2}{a-b} stackrel{AM-GM}{geq}2sqrt{2}$$
answered 11 hours ago
trancelocation
8,0561519
answered 11 hours ago
trancelocation
8,0561519
answered 11 hours ago
trancelocation
8,0561519
answered 11 hours ago
trancelocation
8,0561519
8,0561519
add a comment |
add a comment |
up vote
1
down vote
Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.
answered 11 hours ago
José Carlos Santos
139k18111203
add a comment |
up vote
1
down vote
Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.
answered 11 hours ago
José Carlos Santos
139k18111203
add a comment |
up vote
1
down vote
up vote
1
down vote
Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.
answered 11 hours ago
José Carlos Santos
139k18111203
Consider the map$$begin{array}{rccc}fcolon&(1,infty)&longrightarrow&mathbb R\&a&mapsto&dfrac{a^2+frac1{a^2}}{a-frac1a}.end{array}$$Then$$f'(a)=frac{left(a^2+1right) left(a^4-4 a^2+1right)}{a^2left(a^2-1right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=sqrt{2+sqrt3}$. But $fleft(sqrt{2+sqrt3}right)=2sqrt2$.
answered 11 hours ago
José Carlos Santos
139k18111203
answered 11 hours ago
José Carlos Santos
139k18111203
answered 11 hours ago
José Carlos Santos
139k18111203
answered 11 hours ago
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
up vote
0
down vote
Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.
answered 11 hours ago
TheSimpliFire
11.6k62256
add a comment |
up vote
0
down vote
Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.
answered 11 hours ago
TheSimpliFire
11.6k62256
add a comment |
up vote
0
down vote
up vote
0
down vote
Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.
answered 11 hours ago
TheSimpliFire
11.6k62256
Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$frac{a^2+b^2}{a-b}=frac{a^2+1/a^2}{a-1/a}=frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $ane0,pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-sqrt3,2+sqrt3$$ so $a=sqrt{2+sqrt3}$ and $b=sqrt{2-sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.
answered 11 hours ago
TheSimpliFire
11.6k62256
answered 11 hours ago
TheSimpliFire
11.6k62256
answered 11 hours ago
TheSimpliFire
11.6k62256
answered 11 hours ago
TheSimpliFire
11.6k62256
11.6k62256
add a comment |
add a comment |
up vote
0
down vote
Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$
which is the same as $$(x-sqrt{2})^2geq 0$$
and we are done.
answered 11 hours ago
greedoid
34.3k114488
add a comment |
up vote
0
down vote
Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$
which is the same as $$(x-sqrt{2})^2geq 0$$
and we are done.
answered 11 hours ago
greedoid
34.3k114488
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$
which is the same as $$(x-sqrt{2})^2geq 0$$
and we are done.
answered 11 hours ago
greedoid
34.3k114488
Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 geq 2xsqrt{2}$$
or $$2b^2+2bx+x^2geq 2xsqrt{2}$$
Since $1=ab = b^2+bx$ we have to prove$$2+x^2geq 2xsqrt{2}$$
which is the same as $$(x-sqrt{2})^2geq 0$$
and we are done.
answered 11 hours ago
greedoid
34.3k114488
edited 11 hours ago
answered 11 hours ago
greedoid
34.3k114488
answered 11 hours ago
greedoid
34.3k114488
answered 11 hours ago
greedoid
34.3k114488
34.3k114488
add a comment |
add a comment |
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