1@xyz i couldnt understand your solution as i aint thorough with calculus.
i was unable to solve this one but wait till you see the solution given in the book!!
if a,b,c,d are four non-negative reals and a+b+c+d=1, show that ab+bc+cd \le \frac{1}{4}
-
UP 0 DOWN 0 0 13
13 Answers
1
1applying a.m-g.m inequality three times
(a+b)24 >= ab
(b+c)24 >= bc
(c+d)24 >= cd
adding,
ab + bc + cd <= 14 [ (a+b)2 + (b+c)2 +(c+d)2 ]
r.h.s of the inequality is always >= a+b+c+d if a,b,c,d are all +ve. and this can be easily proved.
so ab + bc + cd <= 14
1soumya
take for example a=b=c=d=\frac{1}{4}
\frac{(a+b)^2+(b+c)^2+(c+a)^2}{4}=\frac{3}{16}
which is less than a+b+c ie \frac{3}{4}
so your last claim isnt true
341Actually there's a complicated proof for the result, but in such cases where the expression is convex in each variable, you only need to evaluate the expression at points where some of the variables are of the form 1/n for some natural number n and the rest are zero.
Here the maximum is attained when c=d=1/2,a=b=0 and it is 1/4.
1prophet sir could you explain me the meaning of "expression is convex in each variable"
1sorry but my statement should have been that
(b+a)2+(b+c)2+(c+d)2 is always <= 1 as a,b,c,d are fractions or any one of them is =1
so ab+bc+cd <= 14
1soumya now it looks ok
here's the elegant proof
put A=a+b+c+d and B=a-b+c-d
so A+B=2(a+c), A-B=2(b+d)
(A+B)(A-B)=4(a+c)(b+d)=A2-B2\leA2=1
so (a+c)(b+d)\le1/4
but (a+c)(b+d)\ge ab+bc+cd
so ab+bc+ac\le1/4
1Yeah,the proof is truly elgant ,judging by the simplicity and obviousness of it.