help with proof question (1 Viewer)

Intilegience · Feb 12, 2024

need help from part b onwards, including a for context.

jimmysmith560 · Feb 12, 2024

Would the following working help with part (c)?

Lith_30 · Feb 12, 2024

ok for part b)
let's just ski[ the first few induction steps

assume that the following statement is true
$(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_k}\right)\geq{}k^2$
hence prove for $n=k+1$ , that is $(a_1+...+a_{k+1})\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)\geq{}(k+1)^2$

$\\LHS=(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)\\\\=(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_{k}}\right)+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+\frac{1}{a_{k+1}}(a_1+...+a_k)\\\\\geq{}k^2+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+\frac{1}{a_{k+1}}(a_1+...+a_k)\\\\\geq{}k^2+\frac{a_{k+1}}{a_1}+\frac{a_1}{a_{k+1}}+...+\frac{a_{k+1}}{a_k}+\frac{a_k}{a_{k+1}}+2$
using part a we can get that
$LHS\geq{}k^2+2k+1=(k+1)^2$

ExtremelyBoredUser · Feb 12, 2024

Lith_30 said:
ok for part b)
let's just ski[ the first few induction steps

assume that the following statement is true
$(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_k}\right)\geq{}k^2$
hence prove for $n=k+1$ , that is $(a_1+...+a_{k+1})\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)\geq{}(k+1)^2$

$\\LHS=(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)\\\\=(a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_{k}}\right)+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+\frac{1}{a_{k+1}}(a_1+...+a_k)\\\\\geq{}k^2+a_{k+1}\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)+\frac{1}{a_{k+1}}(a_1+...+a_k)\\\\\geq{}k^2+\frac{a_{k+1}}{a_1}+\frac{a_1}{a_{k+1}}+...+\frac{a_{k+1}}{a_k}+\frac{a_k}{a_{k+1}}+2$
using part a we can get that
$LHS\geq{}k^2+2k+1=(k+1)^2$

mf after a math2111 lecture going crazy

help with proof question (1 Viewer)

Intilegience

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jimmysmith560

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Lith_30

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ExtremelyBoredUser

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