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Inequality Challange:: Help

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!

Note by Karan Shekhawat
2 years, 2 months ago

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Apply 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, 2 months ago

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@Sandeep Bhardwaj Sir I need Help in one more question . I'am Posting It in next Note . Please Help me there ! Sorry to disturb you . Karan Shekhawat · 2 years, 2 months ago

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@Sandeep Bhardwaj Thanks Sir ! Karan Shekhawat · 2 years, 2 months ago

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

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@Ronak Agarwal Thanks a lot ronak ! Karan Shekhawat · 2 years, 2 months ago

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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, 2 months ago

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@Mvs Saketh 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. Karan Shekhawat · 2 years, 2 months ago

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@Karan Shekhawat 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) Mvs Saketh · 2 years, 2 months ago

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@Mvs Saketh Sorry But shouldn't it be mgh + mv^2/2 = constant. ? Karan Shekhawat · 2 years, 2 months ago

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@Karan Shekhawat i am measuring height from top, so answer is either mgh-mv^2/2 or mv^2/2 - mgh Mvs Saketh · 2 years, 2 months ago

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@Mvs Saketh oh okay . I'am understanding from bottom as refrence . Can You Please add Symbols meaning in your bubble problem. ? Karan Shekhawat · 2 years, 2 months ago

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From A.M. - G.M. inequality {(a + 2b + 3c)/3}^3 >= a2b3c, 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, 2 months ago

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

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@Sudeep Salgia Thanks Sir ! Karan Shekhawat · 2 years, 2 months ago

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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. Mayank Singh · 2 years, 2 months ago

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