1 - x I 1 - y m 1 - zn -1Ix - my - nz for I - m - n - 0 and - N - x - y -z - 0

From part (a) (1+x)^1 gt;1 + lx L.H.S. gt; (1+lx) (1+my) (1+nz) = 1 + (lx+my+nz) + (lmxy+mnyz+nlzx) + lmnxyz gt; 1 + ...

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Byju's Answer

Standard XII

Mathematics

Cube Root of a Complex Number

1 + x I 1 ...

Question

$(1 + x)^{I} (1 + y)^{m} (1 + z)^{n} \geq \frac{1}{I x + m y + n z}$
for I , m , n > 0 and $ϵ$ N , x , y ,z > 0

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Solution

From part (a) $(1 + x)^{1} > 1 + l x$
$L . H . S . > (1 + l x) (1 + m y) (1 + n z)$
$= 1 + (l x + m y + n z) + (l m x y + m n y z + n l z x) + l m n x y z > 1 + l x + m y + n z$

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(1+x)I(1+y)m(1+z)n≥1Ix+my+nz for I , m , n > 0 and ϵ N , x , y ,z > 0

From part (a) (1+x)1>1+lxL.H.S.>(1+lx)(1+my)(1+nz)=1+(lx+my+nz)+(lmxy+mnyz+nlzx)+lmnxyz>1+lx+my+nz

$(1 + x)^{I} (1 + y)^{m} (1 + z)^{n} \geq \frac{1}{I x + m y + n z}$
for I , m , n > 0 and $ϵ$ N , x , y ,z > 0

From part (a) $(1 + x)^{1} > 1 + l x$
$L . H . S . > (1 + l x) (1 + m y) (1 + n z)$
$= 1 + (l x + m y + n z) + (l m x y + m n y z + n l z x) + l m n x y z > 1 + l x + m y + n z$