Show that problem and that equality if fulfilled iff a=b

-1

$begingroup$

If $a, b >0$, show that $$ frac{a}{b^2} + frac{b}{a^2} geq frac{1}{a} +frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$

I tried using elementary consequences of order axioms, but I had trouble applying them. If you can explain this to me, it would be great.

Thanks.

real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality rearrangement-inequality

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edited Jan 11 at 18:39

amWhy

1

asked Apr 26 '18 at 22:56

mariemarie

407

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add a comment | 

-1

$begingroup$

If $a, b >0$, show that $$ frac{a}{b^2} + frac{b}{a^2} geq frac{1}{a} +frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$

I tried using elementary consequences of order axioms, but I had trouble applying them. If you can explain this to me, it would be great.

Thanks.

real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality rearrangement-inequality

share|cite|improve this question

edited Jan 11 at 18:39

amWhy

1

asked Apr 26 '18 at 22:56

mariemarie

407

$endgroup$

add a comment | 

-1

-1

-1

$begingroup$

If $a, b >0$, show that $$ frac{a}{b^2} + frac{b}{a^2} geq frac{1}{a} +frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$

I tried using elementary consequences of order axioms, but I had trouble applying them. If you can explain this to me, it would be great.

Thanks.

real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality rearrangement-inequality

share|cite|improve this question

edited Jan 11 at 18:39

amWhy

1

asked Apr 26 '18 at 22:56

mariemarie

407

$endgroup$

If $a, b >0$, show that $$ frac{a}{b^2} + frac{b}{a^2} geq frac{1}{a} +frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$

I tried using elementary consequences of order axioms, but I had trouble applying them. If you can explain this to me, it would be great.

Thanks.

real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality rearrangement-inequality

real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality rearrangement-inequality

share|cite|improve this question

edited Jan 11 at 18:39

amWhy

1

asked Apr 26 '18 at 22:56

mariemarie

407

share|cite|improve this question

edited Jan 11 at 18:39

amWhy

1

asked Apr 26 '18 at 22:56

mariemarie

407

share|cite|improve this question

share|cite|improve this question

edited Jan 11 at 18:39

amWhy

1

edited Jan 11 at 18:39

amWhy

1

edited Jan 11 at 18:39

amWhy

1

1

asked Apr 26 '18 at 22:56

mariemarie

407

asked Apr 26 '18 at 22:56

mariemarie

407

asked Apr 26 '18 at 22:56

mariemarie

407

407

add a comment | 

add a comment | 

5 Answers
5

active

oldest

votes

2

$begingroup$

Hint:   multiply by $,a^2b^2 gt 0,$, then $;a^3+b^3 ge ab^2+a^2b iff (a-b)(a^2-b^2) ge 0,$.

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

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2

$begingroup$

$$frac13frac{a}{b^2}+frac23frac{b}{a^2}ge sqrt[3]{frac{1}{a^3}}=frac1a$$
by AM-GM. Adding up symmetrically and we are done.

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

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add a comment | 

1

$begingroup$

First,

$$frac{a}{b^2} + frac{b}{a^2}=frac{a^3+b^3}{a^2b^2}ge frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$frac{a^3+b^3}{ab}ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2ge abLeftrightarrow (a-b)^2ge 0$$

And equality iff $a=b$

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

$endgroup$

add a comment | 

1

$begingroup$

Using Cauchy-Schwarz inequality, we have: $left(dfrac{1}{a}+dfrac{1}{b}right)^2= left(dfrac{sqrt{b}}{a}cdot dfrac{1}{sqrt{b}}+dfrac{sqrt{a}}{b}cdot dfrac{1}{sqrt{a}}right)^2le left(dfrac{b}{a^2}+dfrac{a}{b^2}right)left(dfrac{1}{b}+dfrac{1}{a}right)implies dfrac{1}{a}+dfrac{1}{b}le dfrac{a}{b^2}+dfrac{b}{a^2}$

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

$endgroup$

add a comment | 

0

$begingroup$

$(a,b)$ and $left(frac{1}{a^2},frac{1}{b^2}right)$ they are opposite ordered.

Thus, by Rearrangement we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=acdotfrac{1}{b^2}+bcdotfrac{1}{a^2}geq acdotfrac{1}{a^2}+bcdotfrac{1}{b^2}=frac{1}{a}+frac{1}{b}.$$
The equality occurs iff $left(frac{1}{a^2},frac{1}{b^2}right)=left(frac{1}{b^2},frac{1}{a^2}right)$, id est, for $a=b$.

Also, by C-S we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=frac{a^2}{ab^2}+frac{b^2}{ba^2}geqfrac{(a+b)^2}{ab^2+ba^2}=frac{1}{a}+frac{1}{b}.$$

share|cite|improve this answer

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

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5 Answers
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oldest

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5 Answers
5

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oldest

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active

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active

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votes

2

$begingroup$

Hint:   multiply by $,a^2b^2 gt 0,$, then $;a^3+b^3 ge ab^2+a^2b iff (a-b)(a^2-b^2) ge 0,$.

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

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add a comment | 

2

$begingroup$

Hint:   multiply by $,a^2b^2 gt 0,$, then $;a^3+b^3 ge ab^2+a^2b iff (a-b)(a^2-b^2) ge 0,$.

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

$endgroup$

add a comment | 

2

2

2

$begingroup$

Hint:   multiply by $,a^2b^2 gt 0,$, then $;a^3+b^3 ge ab^2+a^2b iff (a-b)(a^2-b^2) ge 0,$.

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

$endgroup$

Hint:   multiply by $,a^2b^2 gt 0,$, then $;a^3+b^3 ge ab^2+a^2b iff (a-b)(a^2-b^2) ge 0,$.

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

share|cite|improve this answer

share|cite|improve this answer

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

answered Apr 26 '18 at 22:59

dxivdxiv

57.8k648101

57.8k648101

add a comment | 

add a comment | 

2

$begingroup$

$$frac13frac{a}{b^2}+frac23frac{b}{a^2}ge sqrt[3]{frac{1}{a^3}}=frac1a$$
by AM-GM. Adding up symmetrically and we are done.

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

$endgroup$

add a comment | 

2

$begingroup$

$$frac13frac{a}{b^2}+frac23frac{b}{a^2}ge sqrt[3]{frac{1}{a^3}}=frac1a$$
by AM-GM. Adding up symmetrically and we are done.

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

$endgroup$

add a comment | 

2

2

2

$begingroup$

$$frac13frac{a}{b^2}+frac23frac{b}{a^2}ge sqrt[3]{frac{1}{a^3}}=frac1a$$
by AM-GM. Adding up symmetrically and we are done.

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

$endgroup$

$$frac13frac{a}{b^2}+frac23frac{b}{a^2}ge sqrt[3]{frac{1}{a^3}}=frac1a$$
by AM-GM. Adding up symmetrically and we are done.

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

share|cite|improve this answer

share|cite|improve this answer

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

answered Apr 26 '18 at 23:06

BAIBAI

1,780518

1,780518

add a comment | 

add a comment | 

1

$begingroup$

First,

$$frac{a}{b^2} + frac{b}{a^2}=frac{a^3+b^3}{a^2b^2}ge frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$frac{a^3+b^3}{ab}ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2ge abLeftrightarrow (a-b)^2ge 0$$

And equality iff $a=b$

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

$endgroup$

add a comment | 

1

$begingroup$

First,

$$frac{a}{b^2} + frac{b}{a^2}=frac{a^3+b^3}{a^2b^2}ge frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$frac{a^3+b^3}{ab}ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2ge abLeftrightarrow (a-b)^2ge 0$$

And equality iff $a=b$

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

$endgroup$

add a comment | 

1

1

1

$begingroup$

First,

$$frac{a}{b^2} + frac{b}{a^2}=frac{a^3+b^3}{a^2b^2}ge frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$frac{a^3+b^3}{ab}ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2ge abLeftrightarrow (a-b)^2ge 0$$

And equality iff $a=b$

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

$endgroup$

First,

$$frac{a}{b^2} + frac{b}{a^2}=frac{a^3+b^3}{a^2b^2}ge frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$frac{a^3+b^3}{ab}ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2ge abLeftrightarrow (a-b)^2ge 0$$

And equality iff $a=b$

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

share|cite|improve this answer

share|cite|improve this answer

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

answered Apr 26 '18 at 23:06

Daniel Constantin PopescuDaniel Constantin Popescu

46616

46616

add a comment | 

add a comment | 

1

$begingroup$

Using Cauchy-Schwarz inequality, we have: $left(dfrac{1}{a}+dfrac{1}{b}right)^2= left(dfrac{sqrt{b}}{a}cdot dfrac{1}{sqrt{b}}+dfrac{sqrt{a}}{b}cdot dfrac{1}{sqrt{a}}right)^2le left(dfrac{b}{a^2}+dfrac{a}{b^2}right)left(dfrac{1}{b}+dfrac{1}{a}right)implies dfrac{1}{a}+dfrac{1}{b}le dfrac{a}{b^2}+dfrac{b}{a^2}$

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

$endgroup$

add a comment | 

1

$begingroup$

Using Cauchy-Schwarz inequality, we have: $left(dfrac{1}{a}+dfrac{1}{b}right)^2= left(dfrac{sqrt{b}}{a}cdot dfrac{1}{sqrt{b}}+dfrac{sqrt{a}}{b}cdot dfrac{1}{sqrt{a}}right)^2le left(dfrac{b}{a^2}+dfrac{a}{b^2}right)left(dfrac{1}{b}+dfrac{1}{a}right)implies dfrac{1}{a}+dfrac{1}{b}le dfrac{a}{b^2}+dfrac{b}{a^2}$

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

$endgroup$

add a comment | 

1

1

1

$begingroup$

Using Cauchy-Schwarz inequality, we have: $left(dfrac{1}{a}+dfrac{1}{b}right)^2= left(dfrac{sqrt{b}}{a}cdot dfrac{1}{sqrt{b}}+dfrac{sqrt{a}}{b}cdot dfrac{1}{sqrt{a}}right)^2le left(dfrac{b}{a^2}+dfrac{a}{b^2}right)left(dfrac{1}{b}+dfrac{1}{a}right)implies dfrac{1}{a}+dfrac{1}{b}le dfrac{a}{b^2}+dfrac{b}{a^2}$

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

$endgroup$

Using Cauchy-Schwarz inequality, we have: $left(dfrac{1}{a}+dfrac{1}{b}right)^2= left(dfrac{sqrt{b}}{a}cdot dfrac{1}{sqrt{b}}+dfrac{sqrt{a}}{b}cdot dfrac{1}{sqrt{a}}right)^2le left(dfrac{b}{a^2}+dfrac{a}{b^2}right)left(dfrac{1}{b}+dfrac{1}{a}right)implies dfrac{1}{a}+dfrac{1}{b}le dfrac{a}{b^2}+dfrac{b}{a^2}$

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

share|cite|improve this answer

share|cite|improve this answer

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

answered Apr 26 '18 at 23:24

DeepSeaDeepSea

71.2k54487

71.2k54487

add a comment | 

add a comment | 

0

$begingroup$

$(a,b)$ and $left(frac{1}{a^2},frac{1}{b^2}right)$ they are opposite ordered.

Thus, by Rearrangement we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=acdotfrac{1}{b^2}+bcdotfrac{1}{a^2}geq acdotfrac{1}{a^2}+bcdotfrac{1}{b^2}=frac{1}{a}+frac{1}{b}.$$
The equality occurs iff $left(frac{1}{a^2},frac{1}{b^2}right)=left(frac{1}{b^2},frac{1}{a^2}right)$, id est, for $a=b$.

Also, by C-S we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=frac{a^2}{ab^2}+frac{b^2}{ba^2}geqfrac{(a+b)^2}{ab^2+ba^2}=frac{1}{a}+frac{1}{b}.$$

share|cite|improve this answer

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

$endgroup$

add a comment | 

0

$begingroup$

$(a,b)$ and $left(frac{1}{a^2},frac{1}{b^2}right)$ they are opposite ordered.

Thus, by Rearrangement we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=acdotfrac{1}{b^2}+bcdotfrac{1}{a^2}geq acdotfrac{1}{a^2}+bcdotfrac{1}{b^2}=frac{1}{a}+frac{1}{b}.$$
The equality occurs iff $left(frac{1}{a^2},frac{1}{b^2}right)=left(frac{1}{b^2},frac{1}{a^2}right)$, id est, for $a=b$.

Also, by C-S we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=frac{a^2}{ab^2}+frac{b^2}{ba^2}geqfrac{(a+b)^2}{ab^2+ba^2}=frac{1}{a}+frac{1}{b}.$$

share|cite|improve this answer

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

$endgroup$

add a comment | 

0

0

0

$begingroup$

$(a,b)$ and $left(frac{1}{a^2},frac{1}{b^2}right)$ they are opposite ordered.

Thus, by Rearrangement we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=acdotfrac{1}{b^2}+bcdotfrac{1}{a^2}geq acdotfrac{1}{a^2}+bcdotfrac{1}{b^2}=frac{1}{a}+frac{1}{b}.$$
The equality occurs iff $left(frac{1}{a^2},frac{1}{b^2}right)=left(frac{1}{b^2},frac{1}{a^2}right)$, id est, for $a=b$.

Also, by C-S we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=frac{a^2}{ab^2}+frac{b^2}{ba^2}geqfrac{(a+b)^2}{ab^2+ba^2}=frac{1}{a}+frac{1}{b}.$$

share|cite|improve this answer

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

$endgroup$

$(a,b)$ and $left(frac{1}{a^2},frac{1}{b^2}right)$ they are opposite ordered.

Thus, by Rearrangement we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=acdotfrac{1}{b^2}+bcdotfrac{1}{a^2}geq acdotfrac{1}{a^2}+bcdotfrac{1}{b^2}=frac{1}{a}+frac{1}{b}.$$
The equality occurs iff $left(frac{1}{a^2},frac{1}{b^2}right)=left(frac{1}{b^2},frac{1}{a^2}right)$, id est, for $a=b$.

Also, by C-S we obtain:
$$frac{a}{b^2}+frac{b}{a^2}=frac{a^2}{ab^2}+frac{b^2}{ba^2}geqfrac{(a+b)^2}{ab^2+ba^2}=frac{1}{a}+frac{1}{b}.$$

share|cite|improve this answer

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

share|cite|improve this answer

share|cite|improve this answer

edited Apr 27 '18 at 1:23

edited Apr 27 '18 at 1:23

edited Apr 27 '18 at 1:23

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

answered Apr 27 '18 at 1:14

Michael RozenbergMichael Rozenberg

103k1891195

103k1891195

add a comment | 

add a comment | 

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