Question

If a, b, c are positive real numbers, prove that :

\[\large{\cfrac{\sqrt{a+b+c} + \sqrt{a}}{b+c} + \cfrac{\sqrt{a+b+c} + \sqrt{b}}{a+c}}\]\[\large{ + \cfrac{\sqrt{a+b+c} + \sqrt{c}}{a+b} \ge \cfrac{9+3\sqrt{3}}{2\sqrt{a+b+c}}}\]

Answer 1

@Mimi_x3 @Miracrown @ganeshie8 @Kainui @ikram002p @iambatman

Any idea friends ?

Answer 2

I am thinkng along this path:

\[\large{\cfrac{\sqrt{a+b+c} + \sqrt{a}}{b+c} = \cfrac{1}{\sqrt{a+b+c} - \sqrt{a}}}\]

Answer 3

THus, our LHS will become:

\[\sum{\cfrac{1}{\sqrt{a+b+c}-\sqrt{a}} \ge \cfrac{9+3\sqrt{3}}{2\sqrt{a+b+c}}}\]

Answer 4

hmm seperate the numerator

Answer 5

In LHS ?

Answer 6

yeah...

Answer 7

we can take common sqrt(a+b+c) after sepratin .....its an idea

Answer 8

Okay:

\[\sqrt{a+b+c}\sum{\cfrac{1}{b+c}} + \sum{\cfrac{\sqrt{a}}{b+c}}\]

Answer 9

Done

So, how should we proceed ?

Answer 10

will see ltr :O

Answer 11

I am trying to be tricky, let's see if we can get anywhere. \[\LARGE a= \alpha ^2 , \ b = \beta ^2, \ c = \gamma ^2\] and \[\Large \bar v = <\alpha, \beta, \gamma>\]

Now I rewrite the equation as: \[\LARGE \frac{ |\bar v | + \alpha } {\beta^2 + \gamma ^2}+\frac{ |\bar v | + \beta } {\alpha^2 + \gamma ^2}+\frac{ |\bar v | + \gamma } {\alpha^2 +\beta^2} \ge \frac{9+3 \sqrt{3}}{2 |\bar v |}\]

Now can we do some sort of magic vector stuff with this?

Answer 12

Actually we can consider the denominators as simply the dot product of the projections of the vectors on each of the three coordinate planes.

Answer 13

Yes! So we can represent each of the three terms summed up here as: \[\LARGE \sum \frac{ \sqrt{ \bar v \cdot \bar v} + ( \bar v \cdot \bar e_n)}{ ( \bar v \cdot \bar v) -( \bar v \cdot \bar e_n)^2}\] where n is the normal vector to the plane. Now look at the bottom! It's just the difference of two squares so it factors nicely.

\[\LARGE \sum \frac{ \sqrt{ \bar v \cdot \bar v} + ( \bar v \cdot \bar e_n)}{[ \sqrt{ \bar v \cdot \bar v} + ( \bar v \cdot \bar e_n)][ \sqrt{( \bar v \cdot \bar v)} -( \bar v \cdot \bar e_n)]}\] becomes \[\LARGE \sum \frac{1}{\sqrt{( \bar v \cdot \bar v)} -( \bar v \cdot \bar e_n)}\]

Answer 14

Now allowing myself to call \[\LARGE \sqrt{ \bar v \cdot \bar v} = l\] since it's just the length and then of course allowing \[\LARGE \bar v \cdot \bar e _n = v_n\] to be the nth component of v, we get: \[\LARGE \frac{1}{l+\alpha}+ \frac{1}{l+\beta}+ \frac{1}{l+\gamma} \ge \frac{9+3\sqrt{3}}{2 l}\] Since a, b, and c are positive and I chose them to be the squares of alpha, beta, and gamma then we will choose alpha, beta, and gamma to be negative numbers. Now this means if we remove an alpha, beta, and gamma from each we will have a smaller value! So: \[\LARGE \frac{1}{l+\alpha}+ \frac{1}{l+\beta}+ \frac{1}{l+\gamma} \ge \frac{3}{l}\] So now let's see, is this a true statement? \[\LARGE \frac{3}{l} \ge \frac{9+3\sqrt{3}}{2 l}\] It isn't, this is a totally false statement so I completely fall on my face since it's obvious that \[\LARGE 2 <1+ \sqrt{3}\] and I have not finished. Good luck everyone else hopefully I did something you can take and fix though!

Answer 15

Nicely tried Kai...

I will try this after a few days (exams time :P)

Answer 16

Ok good luck haha.