Suppose that a,b,c,d > 0. Prove that abc + abd + acd + bcd/(a + b + c) 3 + (a + b + d) 3 + (a + c + d) 3 + (b + c + d) 3 1/27 Solution We know that ((a+b+c)/3) >= (abc)^(1/3) Hence (a+b+c)^3 >= 27abc [AM-GM inequality] similarly (a+b+d)^3 >= 27abd Also (a+c+d)^3 >= 27acd and (b+c+d)^3 >= 27bcd So lets sum it (a+b+c)^3 + (a+b+d)^3 + (a+c+d)^3 + (b+c+d)^3 >= 27(abc + abd + acd + bcd) Hence (abc + abd + acd + bcd)/((a+b+c)^3 + (a+b+d)^3 + (a+c+d)^3 + (b+c+d)^3) <= 1/27 Hence the inequality was reversed.(!! question was wrong) Hence proved.

1. Suppose that a>b>0, that Euclid's algorithm computes gcd(a,b) in n steps, and that a is the
smallest integer with this property (that is, if a'>b'>0 and gcd(a',b') requires n steps, then
a'>=a); show that a and b are consecutive Fibonacci numbers a=fn(plus)2 and b=fn(plus)1.
Solution
Let quotient of a and b is =Q(n)
Remainder of a and b is=R(n)
n is the step.
n-steps sequence for a and b (with quotients q_{n} and remainders r_{n} at nth step) along with
the recursive relationships for Fibonacci numbers:
By Euclid's algorithm
Step (1)
a = bQ{1} + R{1};
can be written as:
F{n+1} = F{n} + F{n-1};
Step (2)
b =R{1}Q{2} + R{2};
F{n} = F{n-1} + F{n-2}
Step (3)
R{1} =R{2}Q{3}+ R{3};
F{n-1} = F{n-2} + F{n-3}
................................................going to n-1 step.
Step (n-1)
R{n-3} =R{n-2}Q{n-1} + R{n-1};
F{3} = F{2} + F{1}
Step ( n)
R{n-2} = R{n-1}Q{n} + 0;

2. F{2} = F{1} + F{0}
Since every Q 1 and R{n-1} 1 = F{1} from Step (n), we have
R{n-2} = R{n-1}Q{n} F{1}*1 + F{0} = F{2}
Going up, Step (n-1) yields
R{n-3} =R{n-2}Q{n-1} +R{n-1} F{2}*1 + F{1} = F{3},
similarly backtracking the steps to Step (1), we'll have finally
a = bQ{1} + R{1} F{n} + F{n-1} = F{n+1}
a>{n+1}
Thus,
a=F{n+2}
similarly,we will get
b+F{n+1}