If a,b,c are all positive real numbers, such that : $\displaystyle{a+2b+3c=6\quad \& \quad { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\ge \cfrac { 16 }{ 9 } }$ Then Find $a+b+c$ .

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A.M.-G.M inequality :$\dfrac{\frac{a}{2}+\frac{a}{2}+b+b+\frac{3c}{2}+\frac{3c}{2}}{6} \geq \left( \frac{a}{2} \cdot \frac{a}{2} \cdot b \cdot b \cdot \frac{3c}{2} \cdot \frac{3c}{2} \right)^{\frac{1}{6}}$

$\implies \dfrac{6}{6} \geq \left( \frac{9}{16} \cdot a^2 \cdot b^2 \cdot c^2 \right)^{\frac{1}{6}}$

$\implies a^2 \cdot b^2 \cdot c^2 \leq \dfrac{16}{9}$

But given condition is : $a^2 \cdot b^2 \cdot c^2 \geq \dfrac{16}{9}$

So Both the conditions will be true simultaneously only if $a^2 \cdot b^2 \cdot c^2 =\dfrac{16}{9}$

which will hold at :

$\dfrac{a}{2}=b=\dfrac{3c}{2}=2$

$\implies a=2,b=1,c=\dfrac{2}{3}$

Hence $\boxed{a+b+c=\dfrac{11}{3}}$

$\ddot \smile$

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Thanks Sir !

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Sir I need Help in one more question . I'am Posting It in next Note . Please Help me there ! Sorry to disturb you .

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Step 1 Change of variables :

Put $a=x , 2b=y , 3c=z$

Our question becomes :

$x+y+z=6 , x^{2}y^{2}z^{2} \ge 64$ , then find $x+\dfrac{y}{2}+\dfrac{z}{3}$

Step -2 Apply $A.M \ge G.M$ to get :

$\frac{x+y+z}{3} \ge \sqrt [ 3 ]{ xyz }$

Put the value of $x+y+z$ and raising to power 6 both sides :

$x^{2}y^{2}z^{2} \le 64$

But $x^{2}y^{2}z^{2} \ge 64$

$\Rightarrow x^{2}y^{2}z^{2}=64$

Now equality occurs when $x=y=z=2$

Hence we get $x+\dfrac{y}{2}+\dfrac{z}{3} = \dfrac{11}{3}$

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Thanks a lot ronak !

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a is 2, b is 1 and c is 2/3

You'll find that between 'a' , '2b' and '3c', the A.M. must be equal to G.M.

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$\displaystyle \frac{a}{2} + \frac{a}{2} + b + b +\frac{3c}{2} + \frac{3c}{2} \geq 6(\frac{9}{16} a^2 b^2 c^2 )^{\frac{1}{6} } \Rightarrow \frac{6}{6} \geq (\frac{9}{16} a^2 b^2 c^2 )^{\frac{1}{6} } \Rightarrow a^2 b^2 c^2 \leq \frac{16}{9}$

Using the AM-GM inequality, we can easily establish the above fact as shown. But we have been given that $\displaystyle a^2 b^2 c^2 \geq \frac{16}{9}$ thereby implying $\displaystyle a^2 b^2 c^2 = \frac{16}{9}$. It also means that it is the equality case in the AM-GM $\displaystyle \Rightarrow \frac{a}{2} = b = \frac{3c}{2}$. Now, trivially, $\displaystyle a+b+c = \frac{11}{3}$.

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Thanks Sir !

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From A.M. - G.M. inequality {(a + 2b + 3c)/3}^3 >= a

2b3c, as abc = 4/3 is the only possibility for three positive numbers a, 2b and 3c such that a = 2b = 3c = 6/2. Sandeep and Sudeep have got it very neat.Log in to reply

i have no new of an approach to help you bro, and everyone has already posted solution, but just an advice, almost whenever you know that there are insufficient equations to solve a problem, be sure that it must be constrained by some inequality, especially when its weird,

Also @KARAN SHEKHAWAT , i have corrected the problem you reported, if you still find it confusing tell me, if not please unreport it, also tell me where you are unable to understand in bubble problem,, i will surely correct it, thankyou :)

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Yes , I Learned solving Inequality problems only aftre joining brilliant.org since My teachers didn't taught me inequalities in deep but thanks for advice. And I deleted my disput , You can check it. And Also I Already Disput in your bubble problem long time ago , but you did not respond it . So I left that question. You can check disput in bubble. But Please state problems parameters clearly , Since I lost My chances after Reporting The Problem.

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I dont get any notifications when someone disputes my problem or mentions me and hence you see, even this note, i found while scrolling, ok sure, i am checking that bubble problem now , thanks

and yes, i have seen your report now,

A conservation law is basically any quantity that remains constant in a process ,like if i say find a conservation law for a ball falling from height 'h' , then the conservation law is mgh-mv^2/2 = k (k=0 , if ball was dropped from rest)

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@Sudeep Salgia @Sandeep Bhardwaj @Rajen Kapur @Deepanshu Gupta @Ronak Agarwal @Mvs Saketh @Pratik Shastri @Azhaghu Roopesh M @megh choksi @Pranjal Jain @Karthik Kannan @brian charlesworth Sir Please Help .

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