Inequality Challange:: Help

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

$\ddot \smile$

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

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.

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

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, $\displaystyle a+b+c = \frac{11}{3}$.

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.

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 :)

@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 .