Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]
Problem
Source: Iran 2005
Tags: inequalities, 3-variable inequality, cyclic inequality, cauchy schwarz
27.08.2005 16:13
Expanding and multiply each side by 2, we obtain\[2\sum\frac{a^2}{b^2}+2\sum\frac{b}{a}\ge6+2\sum\frac{a}{b}\]Now by AM-GM, $\frac{a^2}{b^2}+\frac{a^2}{b^2}+\frac{b}{a}\ge3\frac{a}{b}$. Now it suffices to prove $\sum\frac{a}{b}+\sum\frac{b}{a}\ge6$, which is obvious btw, I think you'd better post inequality in the inequality forum, rather than algebra(though $\text{Inequality}\in\text{Algebra}$ ) [edit]: (I think)post already moved to inequality by mod
27.08.2005 16:37
This problem was given in the British Maths Olympiad 2005 as well.
27.08.2005 18:11
We have from Cauchy Schwarz inequality: $ \left(\frac ab +\frac bc +\frac ca\right)(ab+bc+ca) \geq (a+b+c)^2 $ and $ \left(\frac ab +\frac bc +\frac ca\right)\left(\frac {1}{ab} +\frac {1}{bc} +\frac {1}{ca}\right) \geq \left(\frac 1b +\frac 1c +\frac 1a\right)^2 $ but $ (ab+bc+ca)\left(\frac {1}{ab} +\frac {1}{bc} +\frac{1}{ca}\right)=(a+b+c)\left(\frac 1a +\frac 1b +\frac 1c\right)$ Hence multiplying the above two inequalities we get the desired result.
28.08.2005 04:15
@ British MO: Indeed, see http://www.mathlinks.ro/Forum/viewtopic.php?t=29253 or http://www.mathlinks.ro/Forum/viewtopic.php?t=25355 . darij
29.08.2005 18:21
Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[(\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)\] $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2$=$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ and $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$=$3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $\rightarrow $ we want to prove that : $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ $ \geq$ $3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $rightarrow $ we want to prove this inequality: $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+ \frac{b}{a}+\frac{a}{c}+\frac{c}{b}$ $ \geq$ $3+ \frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{b}{a}+\frac{a}{c}+\frac{c}{b} \geq 3$(?) so now we should prove that $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{(\frac{a}{b}+\frac{c}{a}+\frac{b}{c})^2}{3}$(?) and we have $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ $\geq$ $3$(?) $\rightarrow $ $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ and now we r done.
05.11.2005 14:37
Here is another solution for this inequality By some obvoius calculations the inequality is equivalent with: $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$ Anyway, by substitutions $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ the inequality becomes $(x+y+z)^{2}\geq 3+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ plus condition $xyz=1.$ After all, the inequality is equivalent with this obvious one: $x^{2}+y^{2}+z^{2}+xy+yz+zx\geq 3+x+y+z.$
12.03.2006 19:43
Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[ (\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c) \] let's take it easy: $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}+ \frac{2a}{c} + \frac{2c}{b} + \frac{2b}{a}$ $\geq$ $1+1+1 +$ $\frac ab$ $+ \frac ac$ $+ \frac ba$ $+ \frac{b}{c}+ \frac{c}{b} + \frac{c}{a}$ now: if we solve this , the problem will solved: $\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0;$ $(\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0) b^2a \longleftrightarrow a^3 + b^3 \geq a^2b + b^2a$
12.03.2006 20:24
ok is that right?
24.06.2013 10:37
Put $x=\frac{a}{b}$ & $y,z$ are defined similarly.now multiplying & expanding we have to show $x^2+y^2+z^2+xy+yz+zx\ge 3+x+y+z$.now $xyz=1$ so am-gm gives $\sum_{cyc}xy\ge 3 (*_1)$ now consider $f(t)=t^2-t$ which is convex.so $ f(x)+f(y)+f(z)\ge3f(\frac{x+y+z}{3})$ again $f$ is increasing for $x\ge 0.5$ now am-gm again gives $x+y+z\ge 3$ so $f(\frac{x+y+z}{3})\ge f(1)=0$ which gives $x^2+y^2+z^2\ge x+y+z(*_2)$.adding $(*_1)$ & $(*_2)$ we have the reqd ineq. Sorry for reviving this post but your comments will be helpful
25.06.2013 08:02
jensen wrote: Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[(\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)\] $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2$=$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ and $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$=$3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $\rightarrow $ we want to prove that : $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ $ \geq$ $3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $rightarrow $ we want to prove this inequality: $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+ \frac{b}{a}+\frac{a}{c}+\frac{c}{b}$ $ \geq$ $3+ \frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{b}{a}+\frac{a}{c}+\frac{c}{b} \geq 3$(?) so now we should prove that $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{(\frac{a}{b}+\frac{c}{a}+\frac{b}{c})^2}{3}$(?) and we have $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ $\geq$ $3$(?) $\rightarrow $ $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ and now we r done. $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ please check this step.It is yet not done.
25.06.2013 08:05
pirhoosh wrote: Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[ (\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c) \] let's take it easy: $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}+ \frac{2a}{c} + \frac{2c}{b} + \frac{2b}{a}$ $\geq$ $1+1+1 +$ $\frac ab$ $+ \frac ac$ $+ \frac ba$ $+ \frac{b}{c}+ \frac{c}{b} + \frac{c}{a}$ now: if we solve this , the problem will solved: $\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0;$ $(\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0) b^2a \longleftrightarrow a^3 + b^3 \geq a^2b + b^2a$ yes the lemma is true and it's a very nice solution
25.06.2013 08:15
i think we can use muirhead
26.06.2013 18:00
I don't see any error in Jansen's proof.Please tell me tat, where is the problem ?
01.07.2013 09:45
hello SMOJ can you show your muirhead solution please?
01.07.2013 09:56
@SMOJ the ineq is not symmetric.you can't apply muirhead.
22.12.2013 11:45
In $\triangle ABC$, Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq 3(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})\geq(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]
23.12.2013 06:52
hey sqing the triangle condition is not necessary: $\implies \sum{\frac{a^2}{b^2}}+\sum{\frac{2a}{c}}\geq \sum{\frac{3a}{c}}$ $\implies \sum{\frac{a^2}{b^2}}\geq \sum{\frac{a}{c}}$ which is $\frac{a^2}{b^2}+\frac{b^2}{c^2}\geq \frac{2a}{c}$ abd ading the similar inequalities... check the triangle condition...
23.12.2013 06:53
oo.... maybe for the second ineq. it is important....
23.12.2013 07:03
yes,yes for the second one the triangle condition is indeed important... after the usual Ravi substitution it is coming down to $x^3+y^3+z^2\geq xy^2+yz^2+zx^2$ which is true by re-arrangement
04.12.2015 11:36
My Solution. By subtracting $RHS$ from $LHS$ we get $-$ $\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)$ $=\sum\dfrac{a^2}{b^2}+2\left(\sum\dfrac{a}{c}\right)-\left[3+\left(\sum\dfrac{a}{b}+\sum \dfrac{a}{c}\right)\right]~.$ $=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}\right)+\sum\dfrac{a}{c}-3~.$ $=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}+\dfrac{1}{4}\right)+\sum\dfrac{a}{c}-\dfrac{15}{4}~.$ $=\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}.$ It follows from $AM-GM$ that $\sum\dfrac{a}{c}\ge 3~.~(1)$ And by $CS$ we get $-$ $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(\sum \dfrac{a}{b}-\dfrac{3}{2}\right)^2}{3}\xrightarrow{A\ge G}\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(3-\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}~.~(2)$ By adding $(1),(2)$ we get $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}\geq 3+\dfrac{3}{4}=\dfrac{15}{4}~.$ Or, $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}\geq 0~.$ Or, $\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)\geq 0$ Or, $\boxed{\left(\sum\dfrac{a}{b}\right)^2\geq \left(\sum a\right)\left(\sum\dfrac{1}{a}\right)}~.~\square$
30.01.2016 00:35
Prove, that for $a,b,c>0$ \[(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2 \geq\sqrt{3(a^2+b^2+c^2)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]
04.05.2017 01:37
$x,y,z>0$,prove that \[ \left( {\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{x}} \right) ^{2}- \left( x+y+z \right) \left( {\frac {1}{x}}+{\frac {1}{y}}+{\frac {1} {z}} \right) \geq \left( \frac{3}{2}+{\frac {5}{18}}\,\sqrt {21} \right) \left( {\frac {y}{x}}+{\frac {z}{y}}+{\frac {x}{z}}-3 \right)\]
04.05.2017 01:46
sqing wrote: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=512906#p512906: \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \] just expanding and AM GM
18.02.2018 13:28
Solution of Iran 2005 $\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\iff\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{a}{c}+2\frac{b}{a}+2\frac{c}{b}\ge 3+\frac{a+c}{b}+\frac{a+b}{c}+\frac{c+b}{a}\iff \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} $ Which is true because by AM-GM $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3$ also we have $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ because of Nordic Math Contest 1987 ;D
20.03.2018 20:34
Let $a,b,c$ be positive numbers such that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{49}{4}$$Find all possible values of the expression $$E=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$$. Problem O438 from Mathematical Reflections Issue 1 - 2018 Proposed by Marius Stănean, Zalău, România
22.03.2018 17:42
WolfusA wrote: Let $a,b,c$ be positive numbers such that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{49}{4}$$Find all possible values of the expression $$E=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$$. Problem O438 from Mathematical Reflections Issue 1 - 2018 Proposed by Marius Stănean, Zalău, România I think this problem has been released for a while.(I have seen before) Let $M= \sum_{cyc}\frac{a}{b}$ and $N= \sum_{cyc}\frac{b}{a}$ From the given condition we have: $M+N= \frac{37}{4}$ We want to find the bound of possible values of $E= \sum_{cyc}\frac{a^2}{b^2}=M^2-2N=M^2+2M-\frac{37}{2}$. Consider cubic polynomial $P(t)$ that have 3 positive real roots:$ x=\frac{a}{b} , y=\frac{b}{c} ,z=\frac{c}{a}$ $P(t)=t^3-Mt^2+Nt-1= t^3-Mt^2+(\frac{37}{4}-M)t-1$ Its discriminant = $((-M)(\frac{37}{4}-M))^2+18(-M)(\frac{37}{4}-M)(-1)-4(\frac{37}{4}-M)^3-4(-M)^3(-1)-27(-1)^2 =\frac{1}{16}(4M-17)(M-5)(4M^2-37M-601)$ must be nonnegative. Because $0 \le M \le \frac{37}{4}$ so $4M^2-37M-601=M(4M-37)-601 \le -601 < 0$ and $(4M-17)(M-5) \le 0 , M \in [\frac{17}{4},5]$ We need to check that all $M \in [\frac{17}{4},5]$ cause $P(t)$ to have only positive real roots. Consider $t \le 0$, we have $t^3,-Mt^2,(\frac{37}{4}-M)t \le 0$ and $-1<0$ so $P(t) < 0$ for all $t \le 0$ Hence all roots of $P(t)$ are positive, so $a,b,c>0$ is valid. So $E= M^2+2M-\frac{37}{2} \in [\frac{129}{16},\frac{33}{2}]$ The lowest value $\frac{129}{16}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{4},2,2)$ or permutations. The greatest value $\frac{33}{2}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{2},\frac{1}{2},4)$ or permutations.
23.03.2018 18:29
Nice trick with discriminant.
09.01.2019 18:05
Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] Crux Mathematicorum, Vol. 43(3), March 2017: Let $a, b,c \in R^+$ such that $abc = 1. $Prove that $$a^2b + b^2c + c^2a \geq\sqrt{(a + b + c)(ab + bc + ca)} $$$$\iff\left(\frac ba+\frac cb+\frac ac\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$
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06.07.2019 02:49
Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] $$\left(\frac ab+\frac bc+\frac ca\right)^2= (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)+\left(\frac1{ab}+\frac1{b^2}\right)(a-b)^2+\left(\frac1{bc}+\frac1{c^2}\right)(b-c)^2+\left(\frac1{ca}+\frac1{a^2}\right)(c-a)^2.$$
03.05.2022 11:58
Note that $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^4$ and $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^2(ab+bc+ca)^2 \ge (a+b+c)^4$. Note that we need to prove $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^4 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2(a+b+c)^2$ and we have $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2(ab+bc+ca)^2 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^4(a+b+c)^4$ so we need to prove $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2(a+b+c)^2 \ge (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2(ab+bc+ca)^2$ which is true.
01.02.2023 21:19
one nice!
02.02.2023 01:58
My solution from WOOT: Looking at the quantities, we are motivated to expand both sides of the inequality. We have $$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right)^2 \ge (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$ $$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + 2\left(\frac{b}{a} + \frac{a}{c} + \frac{c}{b} \right) \ge 3 + \frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b}$$ $$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{b}{a} + \frac{a}{c} + \frac{c}{b} \ge 3 + \frac{a}{b} + \frac{c}{a} + \frac{b}{c}$$Substituting $x = \frac{a}{b}, y = \frac{b}{c}$ and $z = \frac{c}{a}$, it suffices to prove that $$x^2 + y^2 + z^2 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3 + x + y + z$$Now note that $$x^2 + \frac{1}{x} \ge 1 + x$$$$\iff (x - 1)^2(x + 1) \ge 0$$which is true. Adding across $x, y, z$ gives us our earlier result, finishing off the problem. Note that $x, y, z > 0$, with equality occurring when $x = y = z = 1$, or $a = b = c$. Our solution is complete.
02.02.2023 04:59
Let $a,b,c $ be positive real numbers . Prove that $$\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$
20.02.2023 16:58
Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] Let $a,b,c>0.$ Prove that $$ \left(\frac ab+\frac bc+\frac ca\right)^2\geq \frac{a+2b}{c}+\frac{b+2c}{a}+\frac{c+2a}{b} \geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$