If y, z>0 and y+z=c , then minimum value of [ (1+ 1 y ) (1+ 1 z ) ]^ 1 / 2 is equal to
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
If \(y, z>0\) and \(y+z=c\), then minimum value of \(\left[\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\right]^{1 / 2}\) is equal to
- \(\frac{c}{2}+1\)
- \(\frac{2}{c}+3\)
- \(1+\frac{2}{c}\)
- \(\frac{c}{2}\)
Answer
(C) 1+ 2 c
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