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Agung

#1 h Oct 23, 2010, 7:07 AM • 2 Y

Y by Adventure10, Mango247

Let $n \ge 3$ be an integer. Consider positive real numbers $a_1, a_2, ... , a_n$ such that $a_1.a_2.a_3...a_n = 1$ . Show that the following inequality holds $\dfrac{a_1 +3}{(a_1+1)^2} + \dfrac{a_2 +3}{(a_2+1)^2} + ... +\dfrac{a_n +3}{(a_n+1)^2} \ge 3$

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#2 h Oct 25, 2010, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Can someone slove this inequality by Jensen inequality ?

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kuing

#3 h Oct 26, 2010, 8:48 AM • 2 Y

Y by Adventure10, Mango247

Agung wrote:

Let $n \ge 3$ be an integer. Consider positive real numbers $a_1, a_2, ... , a_n$ such that $a_1.a_2.a_3...a_n = 1$ . Show that the following inequality holds $\dfrac{a_1 +3}{(a_1+1)^2} + \dfrac{a_2 +3}{(a_2+1)^2} + ... +\dfrac{a_n +3}{(a_n+1)^2} \ge 3$

first we prove that the ineq holds for $n=3$ .
that's equivalent to:
$x,y,z>0,xyz=1,\frac{{x^2 + 3}}{{\left( {x^2 + 1} \right)^2 }} + \frac{{y^2 + 3}}{{\left( {y^2 + 1} \right)^2 }} + \frac{{z^2 + 3}}{{\left( {z^2 + 1} \right)^2 }} \ge 3.$
WLOG assume $x\ge y\ge z$ , then we have $xy\ge1$ , and
$\frac{{x^2 + 3}}{{\left( {x^2 + 1} \right)^2 }} + \frac{{y^2 + 3}}{{\left( {y^2 + 1} \right)^2 }} - 2 \cdot \frac{{xy + 3}}{{\left( {xy + 1} \right)^2 }}$
$= \frac{{(x - y)^2 \left( {3x^4 y^2 + 4x^3 y^3 + 3x^2 y^4 + x^2 + 4xy + y^2 + (x^2 y^2 - 1)(x^2 y^2 + 5) + 4(xy - 1)(x^2 + y^2 )} \right)}}{{\left( {x^2 + 1} \right)^2 \left( {y^2 + 1} \right)^2 \left( {xy + 1} \right)^2 }}$
$\ge0$ ,
let $xy=t\ge1$ , then we just need to prove
$2 \cdot \frac{{t + 3}}{{\left( {t + 1} \right)^2 }} + \frac{{\frac{1}{{t^2 }} + 3}}{{\left( {\frac{1}{{t^2 }} + 1} \right)^2 }} \ge 3$ .
but
$2 \cdot \frac{{t + 3}}{{\left( {t + 1} \right)^2 }} + \frac{{\frac{1}{{t^2 }} + 3}}{{\left( {\frac{1}{{t^2 }} + 1} \right)^2 }} - 3 = \frac{{(t - 1)^2 \left( {2t^3 + 5t^2 + 2t + 3} \right)}}{{(t + 1)^2 \left( {t^2 + 1} \right)^2 }} \ge 0$ ,
so the ineq for $n=3$ is get prove;

next we use induction to prove that the ineq holds for $n\ge3$ .
assume $n=k$ ineq holds, i.e.
$\frac{a_1 +3}{(a_1+1)^2} + \frac{a_2 +3}{(a_2+1)^2} + ... +\frac{a_k +3}{(a_k+1)^2} \ge 3$ ,
then when $n=k+1$ , we have
$\frac{a_1 +3}{(a_1+1)^2} + \frac{a_2 +3}{(a_2+1)^2} + ... + \frac{a_{k-1} +3}{(a_{k-1}+1)^2} +\frac{a_k a_{k+1} +3}{(a_k a_{k+1}+1)^2} \ge 3$ ,
so we just need to prove that
$\frac{a_k +3}{(a_k+1)^2} + \frac{a_{k+1} +3}{(a_{k+1}+1)^2} \ge \frac{a_k a_{k+1} +3}{(a_k a_{k+1}+1)^2}$ ,
in fact
$LHS-RHS=(a_{k}^4 a_{k+1}^3+3 a_{k}^4 a_{k+1}^2+a_{k}^3 a_{k+1}^4+3 a_{k}^3 a_{k+1}^3+7 a_{k}^3 a_{k+1}^2+5 a_{k}^3 a_{k+1}+3 a_{k}^2 a_{k+1}^4+7 a_{k}^2 a_{k+1}^3+7 a_{k}^2 a_{k+1}^2+7 a_{k}^2 a_{k+1}+5 a_{k} a_{k+1}^3+7 a_{k} a_{k+1}^2+3 a_{k} a_{k+1}+a_{k}+a_{k+1}+3)/((a_{k}+1)^2 (a_{k+1}+1)^2 (a_{k} a_{k+1}+1)^2)>0$ ,
so the ineq is true.

equality holds iff when $n=3$ and $x=y=z=1$ , for any $n\ge3$ , when one of $a_k \to 0$ and the other $\to +\infty$ then $LHS \to 3$ .

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#4 h Oct 26, 2010, 1:15 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

Agung wrote:

Let $n \ge 3$ be an integer. Consider positive real numbers $a_1, a_2, ... , a_n$ such that $a_1.a_2.a_3...a_n = 1$ . Show that the following inequality holds $\dfrac{a_1 +3}{(a_1+1)^2} + \dfrac{a_2 +3}{(a_2+1)^2} + ... +\dfrac{a_n +3}{(a_n+1)^2} \ge 3$

first we prove that the ineq holds for $n=3$ .
that's equivalent to:
$x,y,z>0,xyz=1,\frac{{x^2 + 3}}{{\left( {x^2 + 1} \right)^2 }} + \frac{{y^2 + 3}}{{\left( {y^2 + 1} \right)^2 }} + \frac{{z^2 + 3}}{{\left( {z^2 + 1} \right)^2 }} \ge 3.$
WLOG assume $x\ge y\ge z$ , then we have $xy\ge1$ , and
$\frac{{x^2 + 3}}{{\left( {x^2 + 1} \right)^2 }} + \frac{{y^2 + 3}}{{\left( {y^2 + 1} \right)^2 }} - 2 \cdot \frac{{xy + 3}}{{\left( {xy + 1} \right)^2 }}$
$= \frac{{(x - y)^2 \left( {3x^4 y^2 + 4x^3 y^3 + 3x^2 y^4 + x^2 + 4xy + y^2 + (x^2 y^2 - 1)(x^2 y^2 + 5) + 4(xy - 1)(x^2 + y^2 )} \right)}}{{\left( {x^2 + 1} \right)^2 \left( {y^2 + 1} \right)^2 \left( {xy + 1} \right)^2 }}$
$\ge0$ ,
let $xy=t\ge1$ , then we just need to prove
$2 \cdot \frac{{t + 3}}{{\left( {t + 1} \right)^2 }} + \frac{{\frac{1}{{t^2 }} + 3}}{{\left( {\frac{1}{{t^2 }} + 1} \right)^2 }} \ge 3$ .
but
$2 \cdot \frac{{t + 3}}{{\left( {t + 1} \right)^2 }} + \frac{{\frac{1}{{t^2 }} + 3}}{{\left( {\frac{1}{{t^2 }} + 1} \right)^2 }} - 3 = \frac{{(t - 1)^2 \left( {2t^3 + 5t^2 + 2t + 3} \right)}}{{(t + 1)^2 \left( {t^2 + 1} \right)^2 }} \ge 0$ ,
so the ineq for $n=3$ is get prove;

next we use induction to prove that the ineq holds for $n\ge3$ .
assume $n=k$ ineq holds, i.e.
$\frac{a_1 +3}{(a_1+1)^2} + \frac{a_2 +3}{(a_2+1)^2} + ... +\frac{a_k +3}{(a_k+1)^2} \ge 3$ ,
then when $n=k+1$ , we have
$\frac{a_1 +3}{(a_1+1)^2} + \frac{a_2 +3}{(a_2+1)^2} + ... + \frac{a_{k-1} +3}{(a_{k-1}+1)^2} +\frac{a_k a_{k+1} +3}{(a_k a_{k+1}+1)^2} \ge 3$ ,
so we just need to prove that
$\frac{a_k +3}{(a_k+1)^2} + \frac{a_{k+1} +3}{(a_{k+1}+1)^2} \ge \frac{a_k a_{k+1} +3}{(a_k a_{k+1}+1)^2}$ ,
in fact
$LHS-RHS=(a_{k}^4 a_{k+1}^3+3 a_{k}^4 a_{k+1}^2+a_{k}^3 a_{k+1}^4+3 a_{k}^3 a_{k+1}^3+7 a_{k}^3 a_{k+1}^2+5 a_{k}^3 a_{k+1}+3 a_{k}^2 a_{k+1}^4+7 a_{k}^2 a_{k+1}^3+7 a_{k}^2 a_{k+1}^2+7 a_{k}^2 a_{k+1}+5 a_{k} a_{k+1}^3+7 a_{k} a_{k+1}^2+3 a_{k} a_{k+1}+a_{k}+a_{k+1}+3)/((a_{k}+1)^2 (a_{k+1}+1)^2 (a_{k} a_{k+1}+1)^2)>0$ ,
so the ineq is true.

equality holds iff when $n=3$ and $x=y=z=1$ , for any $n\ge3$ , when one of $a_k \to 0$ and the other $\to +\infty$ then $LHS \to 3$ .

thank you kuing,that's a good solution!

maybe we can get something herehttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=77407

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#5 h Oct 27, 2010, 2:27 AM • 2 Y

Y by Adventure10, Mango247

If the product of every triplet $a_{i},a_{j},a_{k}$ is greater than one, then the product of all the $a_{i}$ is greater than one. So there exists a triplet $(a_{i},a_{j},a_{k}) = (a,b,c)$ such that their product is less than one. Then it suffices to prove
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\ge 3$ for $abc \le 1$ .
Since $\frac{p+3}{(p+1)^{2}} \ge \frac{q+3}{(q+1)^{2}}$ for $q \ge p$ , it suffices to prove
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\ge 3$ for $abc =1$ .
Let $a=x^4$ , $b=y^4$ , $c=z^4$ , $x,y,z \ge 0$ , then $xyz=1$ and
$\sum_{cyc}{\left( \frac{a+3}{(a+1)^{2}} - \frac{3}{\left(a^{\frac{6}{4}}+a^{\frac{3}{4}}+1\right)} \right)}$
$=\sum_{cyc}{ \left( \frac{x^4+3}{(x^4+1)^{2}} - \frac{3}{(x^{6}+x^{3}+1)} \right)}$
$=\sum_{cyc}{ \frac{x^3(x-1)^2(x^5+2x^4-x^2+x+3)}{(x^4+1)^{2}(x^6+x^3+1)} \ge 0 }$ since $x^4 +1 \ge 2x^2$ by AM-GM. So it suffices to prove
$\sum_{cyc}{ \frac{1}{(x^6+x^3+1)} \ge 1 }$ with $xyz=1$ .
Let $x^3=\frac{vw}{u^2}$ , $y^3=\frac{wu}{v^2}$ , $z^3=\frac{uv}{w^2}$ . Then
$\sum_{cyc}{ \frac{1}{(x^6+x^3+1)} = \sum_{cyc}{ \frac{u^4}{(u^4+u^2vw+v^2w^2)}}}$
$\ge { \frac{(u^2+v^2+w^2)^2}{(u^4+v^4+w^4+u^2v^2+v^2w^2+w^2u^2+u^2vw+v^2wu+w^2uv)}}$ By Cauchy-Schwarz
So it suffices to prove $(u^2+v^2+w^2)^2 \ge (u^4+v^4+w^4+u^2v^2+v^2w^2+w^2u^2+u^2vw+v^2wu+w^2uv)$
Which is equivalent to $u^2(v-w)^2+v^2(w-u)^2+w^2(u-v)^2 \ge 0$ which is obviously true

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kuing

#6 h Oct 28, 2010, 2:53 AM • 2 Y

Y by Adventure10, Mango247

malcolm wrote:

If the product of every triplet $a_{i},a_{j},a_{k}$ is greater than one, then the product of all the $a_{i}$ is greater than one. So there exists a triplet $(a_{i},a_{j},a_{k}) = (a,b,c)$ such that their product is less than one. Then it suffices to prove
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\ge 3$ for $abc \le 1$ .
Since $\frac{p+3}{(p+1)^{2}} \ge \frac{q+3}{(q+1)^{2}}$ for $q \ge p$ , it suffices to prove
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\ge 3$ for $abc =1$ .
Let $a=x^4$ , $b=y^4$ , $c=z^4$ , $x,y,z \ge 0$ , then $xyz=1$ and
$\sum_{cyc}{( \frac{a+3}{(a+1)^{2}} - \frac{3}{(a^{\frac{6}{4}}+a^{\frac{3}{4}}+1)} )}$
$=\sum_{cyc}{ ( \frac{x^4+3}{(x^4+1)^{2}} - \frac{3}{(x^{6}+x^{3}+1)} )}$
$=\sum_{cyc}{ \frac{x^3(x-1)^2(x^5+2x^4-x^2+x+3)}{(x^4+1)^{2}(x^6+x^3+1)} \ge 0 }$ since $x^4 +1 \ge 2x^2$ by AM-GM. So it suffices to prove
$\sum_{cyc}{ \frac{1}{(x^6+x^3+1)} \ge 1 }$ with $xyz=1$ .
Let $x^3=\frac{vw}{u^2}$ , $y^3=\frac{wu}{v^2}$ , $z^3=\frac{uv}{w^2}$ . Then
$\sum_{cyc}{ \frac{1}{(x^6+x^3+1)} = \sum_{cyc}{ \frac{u^4}{(u^4+u^2vw+v^2w^2)}}}$
$\ge { \frac{(u^2+v^2+w^2)^2}{(u^4+v^4+w^4+u^2v^2+v^2w^2+w^2u^2+u^2vw+v^2wu+w^2uv)}}$ By Cauchy-Schwarz
So it suffices to prove $(u^2+v^2+w^2)^2 \ge (u^4+v^4+w^4+u^2v^2+v^2w^2+w^2u^2+u^2vw+v^2wu+w^2uv)$
Which is equivalent to $u^2(v-w)^2+v^2(w-u)^2+w^2(u-v)^2 \ge 0$ which is obviously true

nice! but how do you think of ${\frac{3}{(a^{\frac{6}{4}}+a^{\frac{3}{4}}+1)} )}$ , how to get this expression ?

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#7 h Oct 28, 2010, 4:55 AM • 1 Y

Y by Adventure10

We know that $\sum_{cyc}{\frac{1}{(x^2+x+1)}\ge 1 }$ with $xyz=1$ is a strong inequality with the same condition, so we guess that
${\left(\frac{a+3}{(a+1)^{2}}-\frac{3}{\left(a^{2r}+a^{r}+1\right)}\right)} \ge 0$
For some $r$ . The expression is zero for any $r$ when $a=1$ , but for it to hold for all $a$ we need to have $1$ as a double root or else it will change sign as a moves from being slightly smaller to slightly bigger than one. So set
$f(a) = \left(\frac{a+3}{(a+1)^{2}}-\frac{3}{\left(a^{2r}+a^{r}+1\right)}\right)$
and we want $f'(1)=0$ . Solving this for $r$ , we get $r=\frac{3}{4}$ , and so we set $a=x^4$ and obtain
$f(a)=\sum_{cyc}{\frac{x^{3}(x-1)^{2}(x^{5}+2x^{4}-x^{2}+x+3)}{(x^{4}+1)^{2}(x^{6}+x^{3}+1)}\ge 0 }$

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kuing

#8 h Oct 28, 2010, 5:05 AM • 1 Y

Y by Adventure10

malcolm wrote:

We know that $\sum_{cyc}{\frac{1}{(x^2+x+1)}\ge 1 }$ with $xyz=1$ is a strong inequality with the same condition, so we guess that
${\left(\frac{a+3}{(a+1)^{2}}-\frac{3}{\left(a^{2r}+a^{r}+1\right)}\right)} \ge 0$
For some $r$ . The expression is zero for any $r$ when $a=1$ , but for it to hold for all $a$ we need to have $1$ as a double root or else it will change sign as a moves from being slightly smaller to slightly bigger than one. So set
$f(a) = \left(\frac{a+3}{(a+1)^{2}}-\frac{3}{\left(a^{2r}+a^{r}+1\right)}\right)$
and we want $f'(1)=0$ . Solving this for $r$ , we get $r=\frac{3}{4}$ , and so we set $a=x^4$ and obtain
$f(a)=\sum_{cyc}{\frac{x^{3}(x-1)^{2}(x^{5}+2x^{4}-x^{2}+x+3)}{(x^{4}+1)^{2}(x^{6}+x^{3}+1)}\ge 0 }$

oh i see, thank you very much, malcolm

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arqady

#9 h Feb 23, 2013, 7:45 PM • 2 Y

Y by Adventure10, Mango247

malcolm wrote:

it suffices to prove
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\ge 3$ for $abc =1$ .

After full expanding we need to prove that $\sum_{cyc}(a^2b+a^2c+3a^2-ab-a-3)\geq0$ ,
which is easy by AM-GM.

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