If n>m,y>n,z>r,(x,y,z>0) such that ā£ā£
ā£ā£xnrmyrmnzā£ā£
ā£ā£=0 then greatest value of 27xyz(xām)(yān)(zār) is
A
-1
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B
8
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C
1
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D
2
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Solution
The correct option is B 8 Given that, ∣∣
∣∣xnrmyrmnz∣∣
∣∣=0(R1→R1−R2R2→R2−R3) ⇒∣∣
∣∣x−mn−y00y−nr−zmnz∣∣
∣∣=0 ⇒(x−m)[z(y−n)−n(r−z)]+m[(n−y)(r−z)]=0 ⇒zz−r+ny−n+mx−m=0 ⇒zz−r+ny−n+1+mx−m+1=2 ⇒zz−r+n+y−ny−n+m+x−mx−m=2 ⇒zz−r+yy−n+xx−m=2
Applying AM -GM Inequality xx−m+yy−n+zz−r3≥(xyz(x−m)(y−n)(z−r))13 xyz(x−m)(y−n)(z−r)≤827
So 27xyz(x−m)(y−n)(z−r)≤8