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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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 \(a+b+c\) .

Please Help Me in solving this Problem.

Thanks!

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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 \(\boxed{a+b+c=\dfrac{11}{3}}\)

\(\ddot \smile\) – Sandeep Bhardwaj · 2 years ago

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– Karan Shekhawat · 2 years ago

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

– Karan Shekhawat · 2 years ago

Thanks Sir !Log in to reply

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 \( x+\dfrac{y}{2}+\dfrac{z}{3} = \dfrac{11}{3}\) – Ronak Agarwal · 2 years ago

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– Karan Shekhawat · 2 years ago

Thanks a lot ronak !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 :) – Mvs Saketh · 2 years ago

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– Karan Shekhawat · 2 years ago

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.Log in to reply

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) – Mvs Saketh · 2 years ago

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– Karan Shekhawat · 2 years ago

Sorry But shouldn't it be mgh + mv^2/2 = constant. ?Log in to reply

– Mvs Saketh · 2 years ago

i am measuring height from top, so answer is either mgh-mv^2/2 or mv^2/2 - mghLog in to reply

– Karan Shekhawat · 2 years ago

oh okay . I'am understanding from bottom as refrence . Can You Please add Symbols meaning in your bubble problem. ?Log in to reply

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. – Rajen Kapur · 2 years agoLog in to reply

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

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, \(\displaystyle a+b+c = \frac{11}{3} \). – Sudeep Salgia · 2 years ago

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– Karan Shekhawat · 2 years ago

Thanks Sir !Log in to reply

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. – Mayank Singh · 2 years ago

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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 . – Karan Shekhawat · 2 years ago

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