Inequality with the condition $abc=1$












2














Let $a,b,c>0$ such that $abc=1$.Prove that
$frac{1}{sqrt{a}}+frac{1}{sqrt{b}}+frac{1}{sqrt{c}}+frac{3}{{a+b+c+1}}ge frac{15}{4}$
My trying: By AM-GM, we need to prove
$frac{3}{{a+b+c+1}} ge frac{3}{4}$
which is not true






inequality uvw







share|cite|improve this question
























  • By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
    – user376343
    10 hours ago
















2














Let $a,b,c>0$ such that $abc=1$.Prove that
$frac{1}{sqrt{a}}+frac{1}{sqrt{b}}+frac{1}{sqrt{c}}+frac{3}{{a+b+c+1}}ge frac{15}{4}$
My trying: By AM-GM, we need to prove
$frac{3}{{a+b+c+1}} ge frac{3}{4}$
which is not true






inequality uvw







share|cite|improve this question
























  • By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
    – user376343
    10 hours ago














2












2








2


1





Let $a,b,c>0$ such that $abc=1$.Prove that
$frac{1}{sqrt{a}}+frac{1}{sqrt{b}}+frac{1}{sqrt{c}}+frac{3}{{a+b+c+1}}ge frac{15}{4}$
My trying: By AM-GM, we need to prove
$frac{3}{{a+b+c+1}} ge frac{3}{4}$
which is not true






inequality uvw







share|cite|improve this question















Let $a,b,c>0$ such that $abc=1$.Prove that
$frac{1}{sqrt{a}}+frac{1}{sqrt{b}}+frac{1}{sqrt{c}}+frac{3}{{a+b+c+1}}ge frac{15}{4}$
My trying: By AM-GM, we need to prove
$frac{3}{{a+b+c+1}} ge frac{3}{4}$
which is not true






inequality uvw




inequality uvw






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









Michael Rozenberg

96.6k1589188




96.6k1589188










asked 11 hours ago









Winter In My Heart

613




613












  • By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
    – user376343
    10 hours ago


















  • By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
    – user376343
    10 hours ago
















By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
– user376343
10 hours ago




By AM-GM is $a+b+c+1geq 4$ which is not good for your inequality.
– user376343
10 hours ago










1 Answer
1






active

oldest

votes


















0














We need to prove that
$$frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{3}{a^2+b^2+c^2+1}geqfrac{15}{4},$$
where $a$, $b$ and $c$ are positives such that $abc=1$.



Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



Thus, we need to prove that
$$frac{3v^2}{w^2}+frac{3w^2}{9u^2-6v^2+w^2}geqfrac{15}{4}$$ or $f(u)geq0,$ where
$$f(u)=frac{v^2}{w^2}+frac{w^2}{9u^2-6v^2+w^2}-frac{5}{4}.$$
But we see that $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $u$,



which happens for equality case of two variables.



Let $b=a$ and $c=frac{1}{a^2}.$



Thus, we need to prove that
$$frac{2}{a}+a^2+frac{3}{2a^2+frac{1}{a^4}+1}geqfrac{15}{4}$$ or
$$(a-1)^2(8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8)geq0.$$
Can you end it now?



For example:
$$8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8=$$
$$=(8a^7-5.6a^4+a)+(16a^6-2a^5-10a^4+4a^3)+(11.6a^4-13a^3-6a^2+8)geq0.$$






share|cite|improve this answer























  • why the last inequality is true. Please help me
    – Winter In My Heart
    7 hours ago










  • @Winter In My Heart I added something. See now.
    – Michael Rozenberg
    3 hours ago











Your Answer





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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0














We need to prove that
$$frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{3}{a^2+b^2+c^2+1}geqfrac{15}{4},$$
where $a$, $b$ and $c$ are positives such that $abc=1$.



Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



Thus, we need to prove that
$$frac{3v^2}{w^2}+frac{3w^2}{9u^2-6v^2+w^2}geqfrac{15}{4}$$ or $f(u)geq0,$ where
$$f(u)=frac{v^2}{w^2}+frac{w^2}{9u^2-6v^2+w^2}-frac{5}{4}.$$
But we see that $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $u$,



which happens for equality case of two variables.



Let $b=a$ and $c=frac{1}{a^2}.$



Thus, we need to prove that
$$frac{2}{a}+a^2+frac{3}{2a^2+frac{1}{a^4}+1}geqfrac{15}{4}$$ or
$$(a-1)^2(8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8)geq0.$$
Can you end it now?



For example:
$$8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8=$$
$$=(8a^7-5.6a^4+a)+(16a^6-2a^5-10a^4+4a^3)+(11.6a^4-13a^3-6a^2+8)geq0.$$






share|cite|improve this answer























  • why the last inequality is true. Please help me
    – Winter In My Heart
    7 hours ago










  • @Winter In My Heart I added something. See now.
    – Michael Rozenberg
    3 hours ago
















0














We need to prove that
$$frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{3}{a^2+b^2+c^2+1}geqfrac{15}{4},$$
where $a$, $b$ and $c$ are positives such that $abc=1$.



Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



Thus, we need to prove that
$$frac{3v^2}{w^2}+frac{3w^2}{9u^2-6v^2+w^2}geqfrac{15}{4}$$ or $f(u)geq0,$ where
$$f(u)=frac{v^2}{w^2}+frac{w^2}{9u^2-6v^2+w^2}-frac{5}{4}.$$
But we see that $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $u$,



which happens for equality case of two variables.



Let $b=a$ and $c=frac{1}{a^2}.$



Thus, we need to prove that
$$frac{2}{a}+a^2+frac{3}{2a^2+frac{1}{a^4}+1}geqfrac{15}{4}$$ or
$$(a-1)^2(8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8)geq0.$$
Can you end it now?



For example:
$$8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8=$$
$$=(8a^7-5.6a^4+a)+(16a^6-2a^5-10a^4+4a^3)+(11.6a^4-13a^3-6a^2+8)geq0.$$






share|cite|improve this answer























  • why the last inequality is true. Please help me
    – Winter In My Heart
    7 hours ago










  • @Winter In My Heart I added something. See now.
    – Michael Rozenberg
    3 hours ago














0












0








0






We need to prove that
$$frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{3}{a^2+b^2+c^2+1}geqfrac{15}{4},$$
where $a$, $b$ and $c$ are positives such that $abc=1$.



Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



Thus, we need to prove that
$$frac{3v^2}{w^2}+frac{3w^2}{9u^2-6v^2+w^2}geqfrac{15}{4}$$ or $f(u)geq0,$ where
$$f(u)=frac{v^2}{w^2}+frac{w^2}{9u^2-6v^2+w^2}-frac{5}{4}.$$
But we see that $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $u$,



which happens for equality case of two variables.



Let $b=a$ and $c=frac{1}{a^2}.$



Thus, we need to prove that
$$frac{2}{a}+a^2+frac{3}{2a^2+frac{1}{a^4}+1}geqfrac{15}{4}$$ or
$$(a-1)^2(8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8)geq0.$$
Can you end it now?



For example:
$$8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8=$$
$$=(8a^7-5.6a^4+a)+(16a^6-2a^5-10a^4+4a^3)+(11.6a^4-13a^3-6a^2+8)geq0.$$






share|cite|improve this answer














We need to prove that
$$frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{3}{a^2+b^2+c^2+1}geqfrac{15}{4},$$
where $a$, $b$ and $c$ are positives such that $abc=1$.



Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



Thus, we need to prove that
$$frac{3v^2}{w^2}+frac{3w^2}{9u^2-6v^2+w^2}geqfrac{15}{4}$$ or $f(u)geq0,$ where
$$f(u)=frac{v^2}{w^2}+frac{w^2}{9u^2-6v^2+w^2}-frac{5}{4}.$$
But we see that $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $u$,



which happens for equality case of two variables.



Let $b=a$ and $c=frac{1}{a^2}.$



Thus, we need to prove that
$$frac{2}{a}+a^2+frac{3}{2a^2+frac{1}{a^4}+1}geqfrac{15}{4}$$ or
$$(a-1)^2(8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8)geq0.$$
Can you end it now?



For example:
$$8a^7+16a^6-2a^5-4a^4-9a^3-6a^2+a+8=$$
$$=(8a^7-5.6a^4+a)+(16a^6-2a^5-10a^4+4a^3)+(11.6a^4-13a^3-6a^2+8)geq0.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 10 hours ago









Michael Rozenberg

96.6k1589188




96.6k1589188












  • why the last inequality is true. Please help me
    – Winter In My Heart
    7 hours ago










  • @Winter In My Heart I added something. See now.
    – Michael Rozenberg
    3 hours ago


















  • why the last inequality is true. Please help me
    – Winter In My Heart
    7 hours ago










  • @Winter In My Heart I added something. See now.
    – Michael Rozenberg
    3 hours ago
















why the last inequality is true. Please help me
– Winter In My Heart
7 hours ago




why the last inequality is true. Please help me
– Winter In My Heart
7 hours ago












@Winter In My Heart I added something. See now.
– Michael Rozenberg
3 hours ago




@Winter In My Heart I added something. See now.
– Michael Rozenberg
3 hours ago


















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