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# Inequality Question : Find the minimum value possible

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Q.)P,Q and R are distinct natural numbers:
If V = $\frac { { P(Q }^{ 2 }+{ R }^{ 2 })+Q(P^{ 2 }+{ R }^{ 2 })+R(P^{ 2 }+{ R }^{ 2 }) }{ PQR }$ , V must be
i)Greater than 5
ii)Greater than 6
iii)Greater than 4
=>V = $\frac { { P(Q }^{ 2 }+{ R }^{ 2 })+Q(P^{ 2 }+{ R }^{ 2 })+R(P^{ 2 }+{ R }^{ 2 }) }{ PQR }$
= $(\frac { Q }{ R } +\frac { R }{ Q } )+(\frac { P }{ R } +\frac { R }{ P } )+(\frac { P }{ Q } +\frac { Q }{ P } )$

From A.M > G.M (Equality is not included because all numbers are distinct)
$\frac { Q }{ R } +\frac { R }{ Q } )$ > 2 as P,Q and R are distinct natural numbers.
Similarly for others.
Hence V > 6.