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please solve this question in the image without substitution method?????

please solve this question in the image without substitution method?????

Grade:12th pass

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
10 years ago
Hello student,
Please find the answer to your question below
(a)First you need to prove that for positive values of x and y
x/y + y/x >= 2
(x - y)² + 2xy = x² + y²
Divide both sides by xy
(x - y)²/(xy) + 2 = x/y + y/x
Reduce both sides by 2
(x - y)²/(xy) = x/y + y/x - 2
Since LHS is squared and x > 0, y > 0, so xy > 0, so LHS nonnegative 0
0 <= x/y + y/x - 2
x/y + y/x >= 2
(a + b)/c + (a + c)/b + (b + c)/a
= a/c + b/c + a/b + c/b + b/a + c/a
= (a/b + b/a) + (a/c + c/a) + (b/c + c/a)
Since a > 0, b > 0, c > 0, they satisfy the above inequality
Thus
(a + b)/c + (a + c)/b + (b + c)/a >= 2 + 2 + 2
(a + b)/c + (a + c)/b + (b + c)/a >= 6
Multiply both sides by abc
(a + b)abc/c + (a + c)abc/b + (b + c)abc/a >= 6abc
(a + b)ab + (a + c)ac + (b + c)bc >= 6abc..............(1)
also we know that as AM>= GM
(a3+b3+c3)/3>=(a3b3c3)1/3
so a3+b3+c3>=3abc...................(2)
Hence from 1 and 2
2(a3+b3+c3)>=bc(b+c)+ca(c+a)+ab(a+b)
(b) from theorem of inequalities
(a0+1+2+b0+1+2+c0+1+2)/3>((a0+b0+c0)/3)((a1+b1+c1)/3)((a2+b2+c2)/3)
Hence (a3+b3+c3)/3>(a+b+c)(a2+b2+c2)/9
(c)As we know AM>HM
(a+b)/2>2ab/a+b
slly(b+c)/2>2bc/b+c
(a+c)/2>2ac/a+c
By adding above 3 equations we get
(bc/b+c)+(ca/c+a)+(ab/a+b)<(a+b+c)/2
(d)As we know that AM>HM
(a+b)/2>2ab/a+b
by reversing lhs and rhs we get
(2/a+b)<(a+b)/2ab
similarly (2/c+a)<(c+a)/2ca
(2/b+c)<(b+c/2bc)
by adding all the above we get
(2/a+b)+(2/b+c)+(2/c+a)<(1/a+1/b+1/c)
Hence a,b,c,d are true

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