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Question

Which of the following hold good?

A

pxqr+qxrp+rxpq>p+q+r, unless p = q = r, or x = 1
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B
[x2+y2+z2x+y+z]x+y+z>xxyyzz>[x+y+z3]x+y+z
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C
[a2+b2a+b]a+b>aabb
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D
aabb>(a+b2)a+b>ab.ba
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Solution

The correct options are
A
pxqr+qxrp+rxpq>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.
(xqr+xqr+p times)
+(xrp+xrp+q times)
+(xpq+xpq+r times)––––––––––––––––––––––––––––––
p+q+r
>{(xqr.xqrp factors).(xrp.xrpq factors).(xpq.xpqr factors)}1(p+q+r)
or pxqr+qxrp+rxpqp+q+r
>{xp(qr)+q(rp)+r(pq)}1(p+q+r)
or pxqr+qxrp+rxpqp+q+r>[x0]1(p+q+r)
or >1
pxqr+qxrp+rxpq>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,xx factors)(y.yf factors)(z.zz factors)]1(x+y+z)
But x+x+x+x times=x.x=x2
and x.x.xx 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.1xx factors)(1y.1yy factors)(1z.1zz 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.1aa factors)×(1b.1b.1bb 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.ab factors)(b.b.ba 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).

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