The correct options are
A
pxq−r+qxr−p+rxp−q>p+q+r, unless p = q = r, or x = 1
B [x2+y2+z2x+y+z]x+y+z>xxyyzz>[x+y+z3]x+y+z
C [a2+b2a+b]a+b>aabb
D aabb>(a+b2)a+b>ab.ba
(a), (b), (c), (d)
Now because A.M. > G.M.
(xq−r+xq−r+……p times)
+(xr−p+xr−p+……q times)
+(xp−q+xp−q+……r times)––––––––––––––––––––––––––––––––
p+q+r
>{(xq−r.xq−r……p factors).(xr−p.xr−p……q factors).(xp−q.xp−q……r factors)}1(p+q+r)
or pxq−r+qxr−p+rxp−qp+q+r
>{xp(q−r)+q(r−p)+r(p−q)}1(p+q+r)
or pxq−r+qxr−p+rxp−qp+q+r>[x0]1(p+q+r)
or >1
∴pxq−r+qxr−p+rxp−q>p+q+r
(b) In case either p=q=r or x=1, inequality becomes equality
Consider x numbers each equal to x, y numbers each equal to y and z numbers each equal to z.
Apply A.M. > G.M. on these x+y+z numbers
(x+x+x+……x times)+(y+y+y+……y times)+(z+z+z+……z times)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
∴ x+y+z
>[(x,x……x factors)(y.y……f factors)(z.z……z factors)]1(x+y+z)
But x+x+x+……x times=x.x=x2
and x.x.x……x factors=xx
∴x2+y2+z2x+y+z>(xxyyzz)1(x+y+z)
or (x2+y2+z2x+y+z)x+y+z>xxyyzz
Above proves the first part. ......... (A)
2nd part:
For 2nd part consider x numbers each equal to 1x,y numbers each equal to 1y,z numbers each equal to 1z. Apply A.M. > G.M. on these x+y+z numbers.
(1x+1x+……x times)+(1y+1y+……y times)+(1z+1z+……z times)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
∴ x+y+z
>[(1x.1x……x factors)(1y.1y……y factors)(1z.1z……z factors)]1(x+y+z)
or 1x.x+1y.y+1z.zx+y+z>[1xx.1yy.1zz]1(x+y+z)
or (3x+y+z)x+y+z>1xxyyzz
Taking reciprocal and changing the sign of inequality, we get
(x+y+z3)x+y+z<xxyyzz
or xxyyzz>(x+y+z3)x+y+z……(B)
Both (A) and (B) prove the required result.
(c) Proceed as in part (b)
Consider a quantities each equal to 1a,b quantities each equal to 1b; then since A.M. > G.M.
[1a+1a+1a+……a times]+[1b+1b+……b times]a+b
>{(1a.1a.1a……a factors)×(1b.1b.1b……b factors)1(a+b)}
or a.1a+b.1ba+b>{1a1b}1(a+b)
or {2a+b}>{1aa.bb}1(a+b)
or {2a+b}a+b>1aabb
Taking reciprocal and hence changing the sign of inequality, we get
{a+b2}a+b<aa.bb
or aa.bb>{a+b2}a+b……(1)
Note: a and b can be integers as well as fractions
(d) Consider b quantities each equal to a and a quantities each equal to b. Then since A.M. > G.M
(a+a+a+……b times)+(b+b+b+……a times)a+b
>[(a.a.a……b factors)(b.b.b……a factors)1(a+b)]
or ab+aba+b>(ab.ba)1(a+b)
or 2aba+b>(ab.ba)1(a+b)
But a+b2 being A.M. of a and b is greater than 2aba+b their H.M.
∴a+b2>2aba+b>[ab.ba]1(a+b)
or [a+b2]a+b>ab.ba……(2)
∴aa.bb>{a+b2}a+b>ab.ba
by (1) of part (c) and (2).