×

$$Q1$$ 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

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

Note by Satyajit Ghosh
10 months, 2 weeks ago

Sort by:

2) Direct using AM GM · 10 months, 2 weeks ago

Can you show?and what about 1st? · 10 months, 2 weeks ago

Apply am gm on a1&a2, a1&a3,...,a1&an,a2&a3........... Add them all · 10 months, 2 weeks ago

What about the 1st one? · 10 months, 2 weeks ago

Thanks I got it · 10 months, 2 weeks ago

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 !!! · 10 months, 2 weeks 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? · 10 months, 2 weeks ago

Did 2nd on my own · 10 months, 2 weeks ago

Did you also use AM-GM? · 10 months, 2 weeks ago

Yeah!! · 10 months, 2 weeks ago