An algebra problem by Wildan Bagus Wicaksono

Using Titu's Lemma or Chauchy Schwarz Engel we get $\begin{aligned} \frac{1}{a} + \frac{4}{b} + \frac{16}{c} &\ge \frac{(1+2+4)^2}{a+b+c} \\ &\ge \frac{7^2}{7} \\ \therefore \frac{1}{a} + \frac{4}{b} + \frac{16}{c} &\ge 7 \end{aligned}$ Equality holds if only if $\displaystyle \frac{1}{a}=\frac{2}{b}=\frac{4}{c}=\frac{1+2+4}{a+b+c}=1$ . So, we get $a=1,b=2,$ and $c=4$ . Then, $21(c-a-b)=21(4-2-1)=21 _{\square}$ .

Let S be equal to our L.H.S. We know that by titu's lemma or extended Cauchy-schwarz inequality

1/a+ 4/b+16/c is greater than or equal to (1+2+4)^2/a+b+c.

Now to find value of the above function let us use the constrain that maximum value of a+b+c is 7.Envoke this. We are almost done here just remains what you call the calculations and a bit of estimation of equalities. S> or equal to 49/7 That implies S> or equal to 7.

Remember Equality in Cauchy -Schwarz Inequality occurs if and only if ai/bi is constant for 2 sets of (a1,a2,a3,.......an) and (b1,b2,b3.........bn). Where ai and bi are nothing but elements of the above sets.

Thus our final step:

1/a=2/b=4/c.

Therefore,a=1,b=2 and c=4.

Remains just a formality 21(c-a-b)

21(4-2-1)

21.Put it in a box,

Thus our final answer is 21.

Sorry for the inconveneince of certain symbols as I am writing by mobile instead of I pad.

Also I would further try a geometrical approach to the question using modern graph theory.If I become successful them I would surely post it.

If my answer helped you then please upvote.