\(Q1\) If \(S_{ n }\) is the sum

\(S_{ n }=1+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +.........+\frac { 1 }{ n } (n>2),\quad prove\quad that\\ n{ (n+1) }^{ \frac { 1 }{ n } }-n<{ S }_{ n }<n-(n-1){ n }^{ \frac { -1 }{ (n-1) } }\)

\(Q2\) If \(a_{1}\), \(a_{2}\), \(a_{3}\),........, \(a_{n}\) are positive real numbers, prove that

\(\sum _{ i<j }^{ }{ \sqrt { { a }_{ i }{ a }_{ j } } } \le \frac { n-1 }{ 2 } ({ a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n })\)

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TopNewest2) Direct using AM GM – Dev Sharma · 1 year, 5 months ago

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– Satyajit Ghosh · 1 year, 5 months ago

Can you show?and what about 1st?Log in to reply

– Dev Sharma · 1 year, 5 months ago

Apply am gm on a1&a2, a1&a3,...,a1&an,a2&a3........... Add them allLog in to reply

– Satyajit Ghosh · 1 year, 5 months ago

What about the 1st one?Log in to reply

– Satyajit Ghosh · 1 year, 5 months ago

Thanks I got itLog in to reply

By using \(A.M-G.M\) we get

\[\sum_{ω=2}^{ω}\sqrt{a_{1}a_{ω}}\leq(\frac{(n-1)a_{1}+\sum_{ω=2}^{ω}a_{ω})}{2}\]

\[\sum_{ω=3}^{ω}\sqrt{a_{2}a_{ω}}\leq(\frac{(n-2)a_{2}+\sum_{ω=3}^{ω}a_{ω})}{2}\] . . . \[\sqrt{a_{ω-1}a_{ω}}\leq\frac{a_{ω-1}+a_{ω}}{2}\]

Adding all above equations we get

\[\sum_{I<J}\sqrt{a_{I}a_{J}}\leq\frac{(n-1)( \sum_{ω=1}^{ω}a_{ω})}{2}\]

Hence proved !!! – Shivam Jadhav · 1 year, 5 months ago

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– Satyajit Ghosh · 1 year, 5 months ago

Thanks but I got it after the hint was dropped by dev. I wanted the hint for 1st question. Can you give me the hints or provide the solution?Log in to reply

@Dev Sharma @Kushagra Sahni @Nihar Mahajan @Surya Prakash @Swapnil Das Adarsh Kumar – Satyajit Ghosh · 1 year, 5 months ago

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– Kushagra Sahni · 1 year, 5 months ago

Did 2nd on my ownLog in to reply

– Satyajit Ghosh · 1 year, 5 months ago

Did you also use AM-GM?Log in to reply

– Kushagra Sahni · 1 year, 5 months ago

Yeah!!Log in to reply

– Satyajit Ghosh · 1 year, 5 months ago

And first question?Log in to reply