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# please solve this question without substituion methodwhich of the following is true (a,b,c>0)? 2(a^3 b^3 +c^3) greater than or equal bc(b+c) +ca(c+a) + ab(a+b) (a^3 + b^3 + c^3)/3 greater than (a+b+c).(a^2+b^2+c^2)/9 bc/(b+c) +ca/(c+a) +ab/(a+b) is less ½ (a+b+c) 2/(b+c) +2/(c+a) + 2/(a+b) is less 1/a +1/b +1/c

Y RAJYALAKSHMI
45 Points
7 years ago
We now Arithmatic mean (AM) > Harmonic mean (HM)
ie., between any two number a,b (a+b)/2 >2ab/(a+b)
=> 2/(a+b)
=> 2/(a+b)
similarily  2/(b+c)
So, 2/(b+c) + 2/(c+a) + 2/(a+b)
=>2/(b+c) + 2/(c+a) + 2/(a+b)  1/a + 1/b + 1/c

SHAIK AASIF AHAMED
7 years ago
Hello student,
(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