Question

Please solve this question I'll thankful to you. Please Please Please.............​

Answer 1

Here , a , b, c, d, and e are real numbers , which are greater than 1 and a + b + c + d + e = 10

To prove -

$\sf{ \bold { \dfrac{ a ^2 }{ c - 1 } + \dfrac{ b ^ 2 }{ d - 1 } + \dfrac{ c ^ 2 }{ e - 1 } + \dfrac{ d ^ 2 } { a - 1 } + \dfrac{ e ^ 2 }{ b - 1 } > 20 }}$

Solution -

This question can be solved quite easily using the Sedrakyan's Inequality .

According to this inequality -

$\sf{ \dfrac{ { a_1 }^ 2 }{ b_1 } + \dfrac{ { a_2 }^ 2 }{ b_2 } + \dfrac{ { a_3 }^ 2 }{ b_3 } + ....... + \dfrac{ { a_n }^ 2 }{ b_n} \geqslant \dfrac{ { a_1 + a_2 + ... + a_n }^2 }{ b_1 + b_2 + ... + b_n } }$

Now , we can use this inequality here ..

$\sf{ \bold { To \: Prove \: - }} \\ \\ \sf{ \implies { \dfrac{ a ^2 }{ c - 1 } + \dfrac{ b ^ 2 }{ d - 1 } + \dfrac{ c ^ 2 }{ e - 1 } + \dfrac{ d ^ 2 } { a - 1 } + \dfrac{ e ^ 2 }{ b - 1 } > 20 }} \\ \\ \sf{ \bold { Theorem \: Used \: - Sedrakyan s \: Inequality }} \\ \\ \sf{ \bold { Proof \: - }} \\ \\ \sf{ \dfrac{ { a }^ 2 }{ c - 1 } + \dfrac{ { b }^ 2 }{ d - 1 } + \dfrac{ { c }^ 2 }{ e - 1 } + \dfrac{ e ^ 2 }{ b - 1 } \geqslant \dfrac{ { ( a + b + c + d + e ) }^2 }{ a + b + c + d + e - 5 } } \\ \\ \sf{ \implies { \dfrac{ { a }^ 2 }{ c - 1 } + \dfrac{ { b }^ 2 }{ d - 1 } + \dfrac{ { c }^ 2 }{ e - 1 } + \dfrac{ e ^ 2 }{ b - 1 } \geqslant \dfrac{ 100 } { 5 } = 20 }} \\ \\ \sf{ \bold { Hence \: Proved }}$