1/a(a+1)+ 1/b(b+1)+1/c(c+1) >= 3/2 if abc=1 and a,b,c>0

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1/a(a+1)+ 1/b(b+1)+1/c(c+1) >= 3/2 if abc=1 and a,b,c>0

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#1 h Apr 28, 2019, 9:28 AM • 1 Y

Y by Adventure10

Prove that if the product of positive numbers $a,b$ and $c$ equals one, then $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}$

This post has been edited 1 time. Last edited by parmenides51, Apr 28, 2019, 9:28 AM
Reason: source typo

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#2 h Apr 28, 2019, 2:18 PM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

Prove that if the product of positive numbers $a,b$ and $c$ equals one, then $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}$

Let $a,b$ and $c$ be positive numbers . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{abc+1}$

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WeakMathemetician

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#3 h Apr 28, 2019, 2:23 PM • 2 Y

Y by Adventure10, Mango247

let $a=x/y,b=y/z,c=z/x$
$\sum_{cyc} \frac{1}{a(a+1)} = \sum_{cyc} \frac{1}{\frac{x}{y}(\frac{x+y}{y})} = \sum_{cyc} \frac{y^2}{x^2+xy}$ $\sum_{cyc} \frac{y^2}{x^2+xy} \ge \frac{(x+y+z)^2}{x^2+y^2+z^2+xy+xz+yz}$ $RHS = 1 + \frac{xy+xz+yz}{x^2+y^2+z^2+xy+xz+yz} \ge 1 + \frac{xy+xz+yz}{2(xy+xz+yz} = 1 + \frac{1}{2} = \frac{3}{2}$
equality iff x=y=z i.e. a=b=c=1

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#4 h Apr 28, 2019, 6:20 PM • 1 Y

Y by Adventure10

WeakMathemetician wrote:

$RHS = 1 + \frac{xy+xz+yz}{x^2+y^2+z^2+xy+xz+yz} \ge 1 + \frac{xy+xz+yz}{2(xy+xz+yz)} = 1 + \frac{1}{2} = \frac{3}{2}$

This part is wrong, since $x^2+y^2+z^2 \geq xy+yz+zx,$ not $\leq.$ But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
$\sum_{cyc} \frac{y^2}{x^2+xy} \geq \frac{(x^2+y^2+z^2)^2}{\sum y^2(x^2+xy)} = \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2+(xy^3+yz^3+zx^3)}\geq \frac{(x^2+y^2+z^2)^2}{\frac{(x^2+y^2+z^2)^2}{3}+\frac{(x^2+y^2+z^2)^2}{3}} = \frac{3}{2}.$

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#5 h Apr 28, 2019, 7:54 PM • 2 Y

Y by Adventure10, Mango247

DoThinh2001 wrote:

WeakMathemetician wrote:

$RHS = 1 + \frac{xy+xz+yz}{x^2+y^2+z^2+xy+xz+yz} \ge 1 + \frac{xy+xz+yz}{2(xy+xz+yz)} = 1 + \frac{1}{2} = \frac{3}{2}$

This part is wrong, since $x^2+y^2+z^2 \geq xy+yz+zx,$ not $\leq.$ But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
$\sum_{cyc} \frac{y^2}{x^2+xy} \geq \frac{(x^2+y^2+z^2)^2}{\sum y^2(x^2+xy)} = \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2+(xy^3+yz^3+zx^3)}\geq \frac{(x^2+y^2+z^2)^2}{\frac{(x^2+y^2+z^2)^2}{3}+\frac{(x^2+y^2+z^2)^2}{3}} = \frac{3}{2}.$

yes you're right sorry - a silly error

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#6 h Apr 28, 2019, 11:34 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b$ and $c$ be positive numbers . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{abc+1}$

Let $a,b$ and $c$ be positive numbers . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\ge \frac{3}{abc+1}$

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#7 h Aug 24, 2019, 1:51 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b$ and $c$ be positive numbers . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge$ $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\ge \frac{3}{abc+1}$

Balkan MO 2006
Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive numbers . Prove that $\frac{1}{a_1(a_1+1)}+\frac{1}{a_2( a_2+1)}+\cdots++\frac{1}{a_n( a_n+1)}\geq$ $\frac{1}{a_1( a_2+1)}+\frac{1}{a_2(a_3+1)}+\cdots+\frac{1}{a_{n-1}( a_n+1)}+\frac{1}{a_n( a_1+1)}\geq\frac{n}{ a_1a_2\cdots a_n+1}.$

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#8 h Dec 7, 2019, 12:08 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b$ and $c$ be positive numbers . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{abc+1}$

Solution of ytChen. Since $a,b,c>0$ , the Cauchy-Schwarz's Inequality gives that
$\begin{align*}\frac 1{a(1+a)}+\frac 1{b(1+b)}+\frac 1{c(1+c)}=\frac{\frac{1}{a^2}}{1+\frac{1}{a}}+\frac{\frac{1}{b^2}}{1+\frac{1}{b}}+\frac{\frac{1}{c^2}}{1+\frac{1}{c}} \ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}. \end{align*}$ It is easy to show that $x\ge y>0\Longrightarrow\frac{x^2}{3+x}\ge\frac{y^2}{3+y}$ . Therefore, the inequality $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}$ implies
$\begin{align*}\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{\frac{9}{\sqrt[3]{(abc)^2}}}{3+\frac{3}{\sqrt[3]{abc}}}=\frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}\ge\frac{3}{1+abc}, \end{align*}$ where the last inequality holds since $x^3+1\ge x(x+1)\ \ (\forall x\in\mathbb{R}^+)$ . We are done. $\blacksquare$

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$ (Lijvzhi)

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#9 h Dec 8, 2019, 2:29 AM • 1 Y

Y by Adventure10

Let $a,b,c,d>0.$ Prove that
$\frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}\geq \frac 4{ 2+abcd}$ Solution of ytChen
Put $(x,y,z,u)=\left(\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}\right)$ . Using the Cauchy-Schwarz's Inequality and AM-GM Inequality, we obtain
$\begin{align*}\frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}=&\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1}+\frac{u^2}{2u+1}\\ \ge&\frac{(x+y+z+u)^2}{2(x+y+z+u)+4}\\ =&\frac{1}{\frac{2}{x+y+z+u}+\frac{4}{(x+y+z+u)^2}}\\ \ge&\frac{1}{\frac{1}{2\sqrt[4]{xyzu}}+\frac{1}{4\sqrt[4]{(xyzu)^2}}}\\ =&\frac{4}{2\sqrt[4]{abcd}+\sqrt[4]{(abcd)^2}}\\ \ge&\frac{4}{2+abcd}, \end{align*}$ where the last inequality holds since $w^4+2-w^2-2w=(w-1)^2(w^2+2w+2)\ge0\ \left(\forall w=\sqrt[4]{abcd}\right)$ . We are done. $\blacksquare$

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#10 h Dec 8, 2019, 4:22 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$ (Lijvzhi)

Solution of Zhangyanzong
Put $(b_1,b_2,\cdots,b_n)=\left(\frac{1}{a_1},\frac{1}{a_2},\cdots,\frac{1}{a_n}\right)$ . Using the C-S and AM-GM , we obtain
$\begin{align*}\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\\=&\frac{b_1^2}{(n-2)b_1+1}+\frac{b_2^2}{(n-2)b_2+1}+\cdots+\frac{b_n^2}{(n-2)b_n+1}\\ \ge&\frac{(b_1+b_2+\cdots+b_n)^2}{(n-2)(b_1+b_2+\cdots+b_n)+n}\\ =&\frac{1}{\frac{n-2}{b_1+b_2+\cdots+b_n}+\frac{n}{(b_1+b_2+\cdots+b_n)^2}}\\ \ge&\frac{1}{\frac{n-2}{n\sqrt[n]{b_1b_2\cdots b_n}}+\frac{1}{n\sqrt[n]{(b_1b_2\cdots b_n)^2}}}\\ =&\frac{n}{(n-2)\sqrt[n]{a_1a_2\cdots a_n}+\sqrt[n]{(a_1a_2\cdots a_n)^2}}\\ \ge&\frac{n}{n-2+a_1a_2\cdots a_n}, \end{align*}$ where the last inequality holds since $w^n+n-2\geq\frac{n}{2}w^2+\frac{n-2}{2}=w^2+\frac{n-2}{2}(w^2+1)\geq w^2+(n-2)w.$
$\forall w=\sqrt[n]{a_1a_2\cdots a_n}$ . We are done. $\blacksquare$

Thank ytChen.

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#11 h Sep 25, 2021, 6:03 AM

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Let $a,b$ and $c$ be positive numbers such that $a+b^2+c^2=1$ . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge 4$ Let $a,b$ and $c$ be positive numbers such that $a+2bc=1$ . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge 4$

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#12 h Sep 27, 2021, 12:19 AM

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sqing wrote:

Let $a,b$ and $c$ be positive numbers such that $a+b^2+c^2=1$ . Prove that $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge 4$

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#13 h May 2, 2024, 4:45 AM

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Let $a,b$ and $c$ be positive numbers such that $ab+bc+ca=3$ . Prove that $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)} \ge \frac{3}{2}$

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