An algebra problem by Wildan Bagus Wicaksono

Algebra Level 1

Given a , b , c R + a,b,c \in \mathbb{R}^+ and 1 a + 4 b + 16 c = 7 \frac{1}{a} + \frac{4}{b} + \frac{16}{c} =7 If a + b + c 7 a+b+c\le 7 , find the value of 21 ( c a b ) 21 (c-a-b) .


The answer is 21.

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2 solutions

Using Titu's Lemma or Chauchy Schwarz Engel we get 1 a + 4 b + 16 c ( 1 + 2 + 4 ) 2 a + b + c 7 2 7 1 a + 4 b + 16 c 7 \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 1 a = 2 b = 4 c = 1 + 2 + 4 a + b + c = 1 \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 , a=1,b=2, and c = 4 c=4 . Then, 21 ( c a b ) = 21 ( 4 2 1 ) = 2 1 21(c-a-b)=21(4-2-1)=21 _{\square} .

D K
Jul 29, 2018

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.

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