Prove that if the product of positive numbers $a,b$ and $c$ equals one, then $\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}$
Problem
Source: Tuymaada 2000 Juniors 4
Tags: algebra, inequalities, three variable inequality
28.04.2019 17:18
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 $$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{abc+1}$$
28.04.2019 17:23
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
28.04.2019 21:20
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: $\sum_{cyc} \frac{y^2}{x^2+xy} \geq \frac{(x^2+y^2+z^2)^2}{\sum y^2(x^2+xy)} = \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2+(xy^3+yz^3+zx^3)}\geq \frac{(x^2+y^2+z^2)^2}{\frac{(x^2+y^2+z^2)^2}{3}+\frac{(x^2+y^2+z^2)^2}{3}} = \frac{3}{2}.$
28.04.2019 22:54
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
29.04.2019 02:34
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 $$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\ge \frac{3}{abc+1}$$
24.08.2019 04:51
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 $$\frac{1}{a_1(a_1+1)}+\frac{1}{a_2( a_2+1)}+\cdots++\frac{1}{a_n( a_n+1)}\geq$$$$\frac{1}{a_1( a_2+1)}+\frac{1}{a_2(a_3+1)}+\cdots+\frac{1}{a_{n-1}( a_n+1)}+\frac{1}{a_n( a_1+1)}\geq\frac{n}{ a_1a_2\cdots a_n+1}.$$
07.12.2019 15:08
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)
08.12.2019 05:29
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. $\blacksquare$
08.12.2019 07:22
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}.$$(Lijvzhi) Solution of Zhangyanzong Put $(b_1,b_2,\cdots,b_n)=\left(\frac{1}{a_1},\frac{1}{a_2},\cdots,\frac{1}{a_n}\right)$. Using the C-S and AM-GM , we obtain \begin{align*}\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\\=&\frac{b_1^2}{(n-2)b_1+1}+\frac{b_2^2}{(n-2)b_2+1}+\cdots+\frac{b_n^2}{(n-2)b_n+1}\\ \ge&\frac{(b_1+b_2+\cdots+b_n)^2}{(n-2)(b_1+b_2+\cdots+b_n)+n}\\ =&\frac{1}{\frac{n-2}{b_1+b_2+\cdots+b_n}+\frac{n}{(b_1+b_2+\cdots+b_n)^2}}\\ \ge&\frac{1}{\frac{n-2}{n\sqrt[n]{b_1b_2\cdots b_n}}+\frac{1}{n\sqrt[n]{(b_1b_2\cdots b_n)^2}}}\\ =&\frac{n}{(n-2)\sqrt[n]{a_1a_2\cdots a_n}+\sqrt[n]{(a_1a_2\cdots a_n)^2}}\\ \ge&\frac{n}{n-2+a_1a_2\cdots a_n}, \end{align*}where the last inequality holds since $w^n+n-2\geq\frac{n}{2}w^2+\frac{n-2}{2}=w^2+\frac{n-2}{2}(w^2+1)\geq w^2+(n-2)w.$ $\forall w=\sqrt[n]{a_1a_2\cdots a_n}$. We are done. $\blacksquare$ Thank ytChen.
25.09.2021 09:03
Let $a,b$ and $c$ be positive numbers such that $a+b^2+c^2=1$ . Prove that$$\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$$
27.09.2021 03:19
sqing wrote: Let $a,b$ and $c$ be positive numbers such that $a+b^2+c^2=1$ . Prove that$$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge 4$$
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02.05.2024 07:45
Let $a,b$ and $c$ be positive numbers such that $ab+bc+ca=3$ . Prove that$$ \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)} \ge \frac{3}{2} $$