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Problem 16

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#1 h May 3, 2010, 6:28 AM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$

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wya

#2 h May 3, 2010, 6:46 AM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

Nguyenhuyen AG wrote:

Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$

By Cauchy Schwarz Inequality, we have
$\sum\frac{1}{a^{4}+b^{3}+c^{2}}$
$=\sum\frac{1+b+c^{2}}{\left(a^{4}+b^{3}+c^{2}\right)\left(1+b+c^{2}\right)}$
$\le\sum\frac{1+b+c^{2}}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}$
$=\frac{3+(a+b+c)+\left(a^{2}+b^{2}+c^{2}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}$

By AM-GM Inequality, we have
$a+b+c\ge3\sqrt[3]{abc}\ge3,$
$a^{2}+b^{2}+c^{2}\ge3\sqrt[3]{a^{2}b^{2}c^{2}}\ge3.$

$\Rightarrow\frac{3+(a+b+c)+\left(a^{2}+b^{2}+c^{2}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}$
$\le\frac{\frac{1}{3}(a+b+c)^{2}+\frac{1}{3}(a+b+c)^{2}+\left(a^{2}+b^{2}+c^{2}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}$
$\le\frac{3}{a^{2}+b^{2}+c^{2}}$
$\le1.$

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zhu rh

#3 h May 3, 2010, 7:14 AM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

my solution
so if $abc>1$ we can let $a$ replaced by $a_{1}$ which is smaller than $a$ and the $LHS$ will be bigger so we just need to prove the situation which $abc=1$ thus according to the CS inequality $(a^{4}+b^{3}+c^{2})(1+b+c^{2})\geqslant (\sum a^{2})$ , $\frac{(1+b+c^{2})}{\sum a^{2}}\geqslant \frac{1}{a^{4}+b^{3}+c^{2}}$ and the other two are simillar so we have $LHS\leqslant \frac{\sum a^{2}+\sum a+3}{(\sum a^{2})^{2}}$
then we just need to prove $\sum a^{4}+2\sum a^{2}b^{2}\geqslant \sum (a^{2}+a+1)$
for $a^{4}+1\geqslant 2a^{2}$
we just need to prove $\sum a^{2}+2\sum a^{2}b^{2}\geqslant a+b+c+6$
then $a^{2}+1\geqslant 2a$
so what we want to show is $\sum a +2\sum a^{2}b^{2}\geqslant 9$ which is a straight applied of AM-GM and $abc=1$

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#4 h May 3, 2010, 7:18 AM • 3 Y

Y by Adventure10, Mango247, cubres

I thing I solved this problem, but I am not sure!!!!!
Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$
$abc\ge1,=>a\ge1$ and if we proof that $\frac{1}{a^{4}+b^{3}+c^{2}}\le1<=>a^4+b^3+c^2\ge1$ , which is true, so I proved this ineq!!!!!!!!!!

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zhu rh

#5 h May 3, 2010, 7:25 AM • 3 Y

Y by Adventure10, Mango247, cubres

arshakus wrote:

I thing I solved this problem, but I am not sure!!!!!
Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$
$abc\ge1,=>a\ge1$ and if we proof that $\frac{1}{a^{4}+b^{3}+c^{2}}\le1<=>a^4+b^3+c^2\ge1$ , which is true, so I proved this ineq!!!!!!!!!!

Oh no you cannot prove it like this what about the other two?

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#6 h May 3, 2010, 7:27 AM • 3 Y

Y by Adventure10, Mango247, cubres

oh yes, now I saw how I was wrong , thank you for explaination!!!!!!

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#7 h May 3, 2010, 7:29 AM • 3 Y

Y by Adventure10, Mango247, cubres

zhu rh wrote:

my solution
so if $abc>1$ we can let $a$ replaced by $a_{1}$ which is smaller than $a$ and the $LHS$ will be bigger so we just need to prove the situation which $abc=1$ thus according to the CS inequality $(a^{4}+b^{3}+c^{2})(1+b+c^{2})\geqslant (\sum a^{2})$ , $\frac{(1+b+c^{2})}{\sum a^{2}}\geqslant \frac{1}{a^{4}+b^{3}+c^{2}}$ and the other two are simillar so we have $LHS\leqslant \frac{\sum a^{2}+\sum a+3}{(\sum a^{2})^{2}}$
then we just need to prove $\sum a^{4}+2\sum a^{2}b^{2}\geqslant \sum (a^{2}+a+1)$
for $a^{4}+1\geqslant 2a^{2}$
we just need to prove $\sum a^{2}+2\sum a^{2}b^{2}\geqslant a+b+c+6$
then $a^{2}+1\geqslant 2a$
so what we want to show is $\sum a +2\sum a^{2}b^{2}\geqslant 9$ which is a straight applied of AM-GM and $abc=1$

what is $LHS$ ??

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zhu rh

#8 h May 3, 2010, 9:55 AM • 3 Y

Y by Adventure10, Mango247, cubres

$LHS$ means $left -hand- side$

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Vasc

#9 h May 5, 2010, 1:39 PM • 2 Y

Y by Adventure10, cubres

The following is nicer:

If $a,b,c$ are positive numbers such that $a+b+c=3$ , then

$\frac 1{a^3+b^2+c}+\frac 1{b^3+c^2+a}+\frac 1{c^3+a^2+b}\le 1.$

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wya

#10 h May 6, 2010, 7:06 AM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

If $a,b,c$ are positive numbers such that $a+b+c=3$ , then

$\frac 1{a^3+b^2+c}+\frac 1{b^3+c^2+a}+\frac 1{c^3+a^2+b}\le 1.$

$\sum\frac{1}{a^{3}+b^{2}+c}$
$=\sum\frac{1+b+c^{2}}{\left(a^{3}+b^{2}+c\right)\left(1+b+c^{2}\right)}$
$\le\sum\frac{1+b+c^{2}}{\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\right)^{2}}$
$=\frac{6+a^{2}+b^{2}+c^{2}}{\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\right)^{2}},$
it suffices to prove
$6+a^{2}+b^{2}+c^{2}\le a^{3}+b^{3}+c^{3}+2a^{\frac{3}{2}}b^{\frac{3}{2}}+2b^{\frac{3}{2}}c^{\frac{3}{2}}+2c^{\frac{3}{2}}a^{\frac{3}{2}}.$

$\sum\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^{2}\left(2a^{2}+4a^{\frac{3}{2}}b^{\frac{1}{2}}-5ab+4a^{\frac{1}{2}}b^{\frac{3}{2}}+2b^{2}\right)\ge0$
$\Rightarrow2\sum a^{\frac{3}{2}}b^{\frac{3}{2}}\ge-\frac{4}{9}\sum a^{3}+\frac{11}{9}\sum\left(a^{2}b+ab^{2}\right),$
it suffices to prove
$\frac{5}{9}\sum a^{3}+\frac{11}{9}\sum\left(a^{2}b+ab^{2}\right)\ge6+a^{2}+b^{2}+c^{2},$
which is equivalent to the obvious inequality
$\frac{2}{9}\sum\left(a^{2}b+bc^{2}\right)\ge\frac{4}{3}abc.$

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Vasc

#11 h May 6, 2010, 7:11 AM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

The following is nicer:

If $a,b,c$ are positive numbers such that $a+b+c=3$ , then

$\frac 1{a^3+b^2+c}+\frac 1{b^3+c^2+a}+\frac 1{c^3+a^2+b}\le 1.$

Your solution is not very nice, Wya. I think tere is a nicer solution.

The following is also nice:

If $a,b,c$ are positive numbers such that $abc=1$ , then

$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

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manlio

#12 h May 6, 2010, 11:44 AM • 2 Y

Y by Adventure10, cubres

Can you please post your proof, Vasc?

Thank you very much.

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#13 h May 6, 2010, 12:10 PM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

Vasc wrote:

The following is nicer:

If $a,b,c$ are positive numbers such that $a+b+c=3$ , then

$\frac 1{a^3+b^2+c}+\frac 1{b^3+c^2+a}+\frac 1{c^3+a^2+b}\le 1.$

Using the Cauchy-Schwarz Inequality, we have
$\frac{1}{a^3+b^2+c}=\frac{a+b^2+c^3}{(a^3+b^2+c)(a+b^2+c^3)} \le \frac{a+b^2+c^3}{(a^2+b^2+c^2)^2}.$
Therefore, it suffices to prove that
$3+t+a^3+b^3+c^3 \le t^2,$
where $t=a^2+b^2+c^2.$

Since $a^3=3a^2-a^2(3-a),$ this inequality can be written as
$t^2-4t-3+[a^2(3-a)+b^2(3-b)+c^2(3-c)] \ge 0.$
Using the Cauchy-Schwarz Inequality again, we get
$a^2(3-a)+b^2(3-b)+c^2(3-c) \ge \frac{[a(3-a)+b(3-b)+c(3-c)]^2}{(3-a)+(3-b)+(3-c)}=\frac{(9-t)^2}{6}.$
And thus, it is enough to check that
$t^2-4t-3+\frac{(9-t)^2}{6} \ge 0,$
which reduces to
$\frac{7}{6}(t-3)^2 \ge 0.$

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#14 h May 6, 2010, 12:25 PM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

wya wrote:

it suffices to prove
$6+a^{2}+b^{2}+c^{2}\le a^{3}+b^{3}+c^{3}+2a^{\frac{3}{2}}b^{\frac{3}{2}}+2b^{\frac{3}{2}}c^{\frac{3}{2}}+2c^{\frac{3}{2}}a^{\frac{3}{2}}.$

This can be proved simpler:

Using the AM-GM Inequality, we have
$(ab)^{3/2}+(ab)^{3/2}+1 \ge 3ab.$
Therefore,
$2[(ab)^{3/2}+(bc)^{3/2}+(ca)^{3/2}] \ge 3(ab+bc+ca)-3.$
It suffices to prove that
$a^3+b^3+c^3+3(ab+bc+ca) \ge a^2+b^2+c^2+9,$
or
$\begin{aligned} 3(a^3+b^3+c^3)&+3(a+b+c)(ab+bc+ca) \ge \\ &\ge (a+b+c)(a^2+b^2+c^2)+(a+b+c)^3.\end{aligned}$
After expanding, we get the third degree Schur's Inequality
$a^3+b^3+c^3+3abc \ge ab(a+b)+bc(b+c)+ca(c+a).$

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#15 h May 6, 2010, 12:58 PM • 3 Y

Y by Adventure10, Mango247, cubres

Nguyenhuyen AG wrote:

Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$

I just remember a similar one with the same conditon:
$\frac{1}{a+b^{2008}+c^{2009}}+\frac{1}{b+c^{2008}+a^{2009}}+\frac{1}{c+a^{2008}+b^{2009}}\leq 1$

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wya

#16 h May 6, 2010, 1:25 PM • 2 Y

Y by Adventure10, cubres

can_hang2007 wrote:

wya wrote:

it suffices to prove
$6+a^{2}+b^{2}+c^{2}\le a^{3}+b^{3}+c^{3}+2a^{\frac{3}{2}}b^{\frac{3}{2}}+2b^{\frac{3}{2}}c^{\frac{3}{2}}+2c^{\frac{3}{2}}a^{\frac{3}{2}}.$

This can be proved simpler:

Using the AM-GM Inequality, we have
$(ab)^{3/2}+(ab)^{3/2}+1 \ge 3ab.$
Therefore,
$2[(ab)^{3/2}+(bc)^{3/2}+(ca)^{3/2}] \ge 3(ab+bc+ca)-3.$
It suffices to prove that
$a^3+b^3+c^3+3(ab+bc+ca) \ge a^2+b^2+c^2+9,$
or
$\begin{aligned} 3(a^3+b^3+c^3)&+3(a+b+c)(ab+bc+ca) \ge \\ &\ge (a+b+c)(a^2+b^2+c^2)+(a+b+c)^3.\end{aligned}$
After expanding, we get the third degree Schur's Inequality
$a^3+b^3+c^3+3abc \ge ab(a+b)+bc(b+c)+ca(c+a).$

Very nice! can_hang2007.
Thank you for make it nicer.

P.s: Your proof in #13 is also nice.

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Vasc

#17 h May 7, 2010, 8:17 PM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

If $a,b,c$ are positive numbers such that $abc=1$ , then

$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

By expanding, we need to prove that
$\sum (\frac b{a^2}+\frac {a^2}{b}) \ge \sum a(a+1)$ .

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#18 h May 7, 2010, 11:26 PM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

The following is also nice:

If $a,b,c$ are positive numbers such that $abc=1$ , then

$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

This is Problem 27 - section 1 - chapter 2 in the book I sent you, Mr Vasile.

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Vasc

#19 h May 8, 2010, 7:23 AM • 3 Y

Y by Adventure10, Mango247, cubres

can_hang2007 wrote:

Vasc wrote:

The following is also nice:

If $a,b,c$ are positive numbers such that $abc=1$ , then

$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

This is Problem 27 - section 1 - chapter 2 in the book I sent you, Mr Vasile.

Indeed, I posted this inequality on this Forum some time ago. Your proof in book without expanding is very nice. However, the inequality above is interesting:
$\sum (\frac b{a^2}+\frac {a^2}{b}) \ge \sum a(a+1)$ .
Moreover, for $abc=1$ , the following sharper inequality holds
$\sum\frac {a^2}{b}+\sum \frac b{a^2} \ge 2\sum a^2.$
I wait nice proofs for these last inequalities.

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#20 h May 8, 2010, 2:33 PM • 5 Y

Y by huyvietnam, Adventure10, Mango247, cubres, and 1 other user

Vasc wrote:

Moreover, for $abc=1$ , the following sharper inequality holds
$\sum\frac {a^2}{b}+\sum \frac b{a^2} \ge 2\sum a^2.$
I wait nice proofs for these last inequalities.

Using the AM-GM Inequality, we have
$\begin{aligned} 20\cdot \frac{a^2}{b}+1\cdot \frac{c^2}{a} +4\cdot \frac{c}{b^2}+17\cdot \frac{a}{c^2} &\ge 42\sqrt[42]{\left(\frac{a^2}{b}\right)^{20}\left(\frac{c^2}{a}\right)^1\left(\frac{c}{b^2}\right)^4\left(\frac{a}{c^2}\right)^{17}}\\ &=42\sqrt[42]{\frac{a^{56}}{(bc)^{28}}}=42\sqrt[42]{a^{84}}=42a^2.\end{aligned}$
Adding this and its cyclic inequalities, and then dividing the resulting inequality by $21,$ we get the desired result.

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Vasc

#21 h May 8, 2010, 9:25 PM • 2 Y

Y by Adventure10, cubres

can_hang2007 wrote:

Vasc wrote:

Moreover, for $abc=1$ , the following sharper inequality holds
$\sum\frac {a^2}{b}+\sum \frac b{a^2} \ge 2\sum a^2.$
I wait nice proofs for these last inequalities.

Using the AM-GM Inequality, we have
$\begin{aligned} 20\cdot \frac{a^2}{b}+1\cdot \frac{c^2}{a} +4\cdot \frac{c}{b^2}+17\cdot \frac{a}{c^2} &\ge 42\sqrt[42]{\left(\frac{a^2}{b}\right)^{20}\left(\frac{c^2}{a}\right)^1\left(\frac{c}{b^2}\right)^4\left(\frac{a}{c^2}\right)^{17}}\\ &=42\sqrt[42]{\frac{a^{56}}{(bc)^{28}}}=42\sqrt[42]{a^{84}}=42a^2.\end{aligned}$
Adding this and its cyclic inequalities, and then dividing the resulting inequality by $21,$ we get the desired result.

Very nice proof, Can!

The following generalization is valid.
Let $a,b,c$ be positive real numbers such that $abc=1$ . If $k\ge 1$ , then

$(k-1)\sum\frac {a^k}{b}+\sum \frac b{a^k} \ge k\sum a^k$ .

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#22 h May 11, 2010, 9:55 AM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

Vasc wrote:

Very nice proof, Can!

The following generalization is valid.
Let $a,b,c$ be positive real numbers such that $abc=1$ . If $k\ge 1$ , then

$(k-1)\sum\frac {a^k}{b}+\sum \frac b{a^k} \ge k\sum a^k$ .

I used the same method and got this solution.

Attachments:

deeee.pdf (107kb)

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#23 h May 11, 2010, 10:31 AM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

The following is nicer:

If $a,b,c$ are positive numbers such that $a+b+c=3$ , then

$\frac 1{a^3+b^2+c}+\frac 1{b^3+c^2+a}+\frac 1{c^3+a^2+b}\le 1.$

Using the inequality Cauchy-Schwarz we have
$(a^2+b^2+c^2)^2\le (a^3+b^2+c)(a+b^2+c^3)$
$\Rightarrow \frac{1}{a^3+b^2+c}\le\frac{a+b^2+c^3}{(a^2+b^2+c^2)^2}$
So
$\sum \frac{1}{a^3+b^2+c}\le\frac{a^3+b^3+c^3+a^2+b^2+c^2+3}{(a^2+b^2+c^2)^2}$
We prove that
$(a^2+b^2+c^2)^2 \ge a^3+b^3+c^3+a^2+b^2+c^2+3$
Let $q=ab+bc+ca,r=abc$ , we have
${3=a+b+c \Rightarrow \{\begin{matrix} 3\ge q\\ 1\ge r \end{matrix}}$
and
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$(a+b+c)^3=a^3+b^3+c^3+3[(a+b+c)(ab+bc+ca)-abc]$
Inequality is rewritten as follows
$4(q-3)^2+3(1-r)+(3-q) \ge 0$
The end poof

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Vasc

#24 h May 12, 2010, 5:34 AM • 3 Y

Y by Adventure10, Mango247, cubres

can_hang2007 wrote:

Vasc wrote:

Very nice proof, Can!

The following generalization is valid.
Let $a,b,c$ be positive real numbers such that $abc=1$ . If $k\ge 1$ , then

$(k-1)\sum\frac {a^k}{b}+\sum \frac b{a^k} \ge k\sum a^k$ .

I used the same method and got this solution.

Very nice proof, Can.
In the same conditions we have also:

$(k^2-1)\sum\frac {a^k}{b}+(k+2)\sum \frac b{a^k} \ge (k^2+k+1)\sum a^k$ .

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hctb00

#25 h Aug 17, 2014, 4:52 AM • 3 Y

Y by Adventure10, Mango247, cubres

$a,b,c>0,abc=1$
1: $p,q>0$ , $\frac{2}{3}\min\{p,q\}\ge\frac{1}{3}\max\{p,q\}$ ,then: $\frac{1}{a^p+b^q+1}+\frac{1}{b^p+c^q+1}+\frac{1}{c^p+a^q+1}\le1$
2: $p,q,r>0,\frac{2}{3}\min\{p,q,r\}\ge\frac{1}{3}\max\{p,q,r\}$ ,then:
$\frac{1}{a^p+b^q+c^r}+\frac{1}{b^p+c^q+a^r}+\frac{1}{c^p+a^q+b^r}\le1$

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41803 posts

sqing

#26 h Dec 25, 2015, 12:36 PM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

can_hang2007 wrote:

Vasc wrote:

The following is also nice:
If $a,b,c$ are positive numbers such that $abc=1$ , then
$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

This is Problem 27 - section 1 - chapter 2 in the book I sent you, Mr Vasile.

Indeed, I posted this inequality on this Forum some time ago. *

http://www.artofproblemsolving.com/community/c6h183001p1005736

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john10

#27 h Dec 25, 2015, 3:10 PM • 3 Y

Y by Adventure10, Mango247, cubres

zhu rh wrote:

my solution
so if $abc>1$ we can let $a$ replaced by $a_{1}$ which is smaller than $a$ and the $LHS$ will be bigger so we just need to prove the situation which $abc=1$ thus according to the CS inequality $(a^{4}+b^{3}+c^{2})(1+b+c^{2})\geqslant (\sum a^{2})$ , $\frac{(1+b+c^{2})}{\sum a^{2}}\geqslant \frac{1}{a^{4}+b^{3}+c^{2}}$ and the other two are simillar so we have $LHS\leqslant \frac{\sum a^{2}+\sum a+3}{(\sum a^{2})^{2}}$
then we just need to prove $\sum a^{4}+2\sum a^{2}b^{2}\geqslant \sum (a^{2}+a+1)$
for $a^{4}+1\geqslant 2a^{2}$
we just need to prove $\sum a^{2}+2\sum a^{2}b^{2}\geqslant a+b+c+6$
then $a^{2}+1\geqslant 2a$
so what we want to show is $\sum a +2\sum a^{2}b^{2}\geqslant 9$ which is a straight applied of AM-GM and $abc=1$

Can you verify please if this solution is true ... to prove it true for abc>=1 he proves it just for abc=1 !

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sqing

#28 h Jan 11, 2016, 7:42 AM • 2 Y

Y by Adventure10, cubres

Vasc wrote:

The following is also nice:
If $a,b,c$ are positive numbers such that $abc=1$ , then
$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

If $a,b,c$ are positive numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ , then
$\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\le 1$

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sqing

#29 h Apr 29, 2016, 7:29 AM • 2 Y

Y by Adventure10, cubres

If $a,b,c$ are positive numbers .Then $\frac{1}{a^2+ b+1}+\frac{1}{b^2+ c+1}+\frac{1}{c^2 +a+1}\leq\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$

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arqady

#30 h Apr 29, 2016, 7:46 AM • 3 Y

Y by Adventure10, Mango247, cubres

sqing wrote:

If $a,b,c$ are positive numbers .Then $\frac{1}{a^2+ b+1}+\frac{1}{b^2+ c+1}+\frac{1}{c^2 +a+1}\leq\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$

By AM-GM and C-S $\sum_{cyc}\frac{1}{a^2+ b+1}\leq\sum_{cyc}\frac{1}{2a+b}\leq\frac{1}{9}\sum_{cyc}\left(\frac{2^2}{2a}+\frac{1^2}{b}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ .

This post has been edited 1 time. Last edited by arqady, Apr 29, 2016, 7:46 AM

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sqing

#32 h Feb 13, 2017, 5:52 AM • 3 Y

Y by Adventure10, Mango247, cubres

Vasc wrote:

*
The following is also nice:

If $a,b,c$ are positive numbers such that $abc=1$ , then

$\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\le 1.$

here
Let $a, b, c>0$ , and $abc=1$ . Prove that
$\frac{3}{ab+bc+ca} \le \frac{1}{a^2+b+1}+ \frac{1}{b^2+c+1}+ \frac{1}{c^2+a+1}\leq 1$

This post has been edited 1 time. Last edited by sqing, Feb 13, 2017, 5:56 AM

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sqing

#33 h Jan 29, 2019, 1:36 AM • 3 Y

Y by Adventure10, Mango247, cubres

Let $a,b,c$ are positive numbers such that $a+b+c=3 .$ Prove that
$\frac{1}{a^2+b+c} +\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b} \le1$ Marin Chirciu:
Let $a,b,c$ are positive numbers such that $a+b+c=3.$ Prove that $\frac{1}{a^2+k(b+c)} +\frac{1}{b^2+k(c+a)}+\frac{1}{c^2+k(a+b)} \le\frac{3}{2k+1}.$ where $1\leq k\leq 2.$
Let be $n\in N, n\geq 2.$ Prove that for $a_1,a_2,\cdots,a_n >0 ,$ with $a_1+a_2+\cdots+a_n= n$ holds the following inequality:
$\frac{1}{a^2_1+a_2+a_3+\cdots+a_n}+\frac{1}{a^2_2+a_3+a_4+\cdots+a_n+a_1}+\cdots+\frac{1}{a^2_n+a_1+a_2+\cdots+a_{n-1}}\leq 1.$ Let be $n\in N, n\geq 2.$ Prove that for $a_1,a_2,\cdots,a_n >0 ,$ with $a_1+a_2+\cdots+a_n= n$ holds the following inequality:
$\frac{1}{a^2_1+k(a_2+a_3+\cdots+a_n)}+\frac{1}{a^2_2+k(a_3+a_4+\cdots+a_n+a_1)}+\cdots+\frac{1}{a^2_n+k(a_1+a_2+\cdots+a_{n-1})}\leq \frac{n}{(n-1)k+1}.$ where $1\leq k\leq 2.$

For $a, b, c>0, a+b+c=3 .$ Prove that
$\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}$
Proof of ytChen:
The Cauchy-Schwarz's Inequality gives $(x+y+z+1)^2\le(x^2+y+z+1)(1+y+z+1)$ for any positive real numbers $x,y,z$ , and thus
$\begin{align*}&\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\\ \leq&\frac{2+b+c}{(a+b+c+1)^2}+\frac{2+c+a}{(a+b+c+1)^2}+\frac{2+a+b}{(a+b+c+1)^2}\\ =&\frac{6+2(a+b+c)}{(a+b+c+1)^2}=\frac{3}{4}. \end{align*}$ As required. $\blacksquare$
https://artofproblemsolving.com/community/c6h1838005p12339974

This post has been edited 2 times. Last edited by sqing, May 16, 2019, 12:40 PM

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ytChen

#34 h Jun 19, 2019, 9:43 PM • 3 Y

Y by Adventure10, Mango247, cubres

sqing wrote:

Let $a,b,c$ are positive numbers such that $a+b+c=3 .$ Prove that
$\frac{1}{a^2+b+c} +\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b} \le1$

Solution. The Cauchy-Schwarz's Inequality gives $(x+y+z)^2\le(x^2+y+z)(1+y+z)$ for any $x,y,z\in\mathbb{R}^+$ , hence
$\begin{align*}&\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\\ \le&\frac{1+b+c}{(a+b+c)^2}+\frac{1+c+a}{(a+b+c)^2}+\frac{1+a+b}{(a+b+c)^2}\\ =&\frac{3+2(a+b+c)}{(a+b+c)^2}=1\,(\because a+b+c=3). \end{align*}$ As required. $\blacksquare$

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sqing

#35 h Jun 19, 2019, 11:24 PM • 2 Y

Y by Adventure10, cubres

Thanks.
Let $a,b,c$ are positive numbers such that $a+b+c=3$ or $abc=1.$ Prove that
$\frac{a}{a^2+b+c} +\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b} \le1 .$

sqing wrote:

Let $a,b,c$ are positive numbers such that $a+b+c=3 .$ Prove that
$\frac{1}{a^2+b+c} +\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b} \le1$

Let $a,b,c$ are positive real numbers. Prove that
$\left[4(a+b+c)-3\right]\left(\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\right) \leqslant 9.$ h

This post has been edited 2 times. Last edited by sqing, Oct 13, 2022, 6:04 AM

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ytChen

#36 h Jun 20, 2019, 6:47 PM • 3 Y

Y by Adventure10, Mango247, cubres

sqing wrote:

Let $a,b,c$ are positive numbers such that $a+b+c=3$ or $abc=1.$ Prove that
$\frac{a}{a^2+b+c} +\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b} \le1 .$

Solution. First observe that $abc=1$ together with $a,b,c>0$ implies $a+b+c\ge3$ . By the Cauchy-Schwarz's Inequality we get $(x+y+z)^2\le(x^2+y+z)(1+y+z)$ for any $x,y,z\in\mathbb{R}^+$ , from which it follows
$\begin{align*}&\frac{a}{a^2+b+c}+\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b}\\ \le&\frac{a(1+b+c)}{(a+b+c)^2}+\frac{b(1+c+a)}{(a+b+c)^2}+\frac{c(1+a+b)}{(a+b+c)^2}\\ =&\frac{1}{a+b+c}+\frac{2(ab+bc+ca)}{(a+b+c)^2}\\ \le&\frac{1}{3}+\frac{2}{3}=1. \end{align*}$ The last inequality holds since $a+b+c\ge3$ and $(a+b+c)^2\ge3(ab+bc+ca)$ . As required. $\blacksquare$

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sqing

#37 h Oct 31, 2021, 2:53 AM • 1 Y

Y by cubres

Let $a,b,c\geq 0$ and $a^2+2bc= 1.$ Prove that
$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1} \leq \frac{7}{3}$ $\frac 1{a^2+b+1}+\frac 1{b^2+c+1}+\frac 1{c^2+a+1}\leq 2$ $\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1}+\frac{8}{(1+a)(1+b)(1+c)} \leq \frac{19}{3}$ $\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}+\frac{8}{(1+a)(1+b)(1+c)}\leq 6$

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sqing

#38 h Mar 17, 2023, 2:09 PM • 1 Y

Y by cubres

Let $a,b,c$ are positive numbers such that $a+b+c=3.$ Prove that
$\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \le\frac{3}{ab+bc+ca}$

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#39 h Apr 13, 2025, 1:45 AM • 1 Y

Y by cubres

Generalization:
$\forall a,b,c\in\mathbb{R}^+,abc=1$ ,there's $\sum\limits_{cyc}\frac{1}{a+b^x+c^y}\leq1\ (1\leq x\leq y\leq 2x)$

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