Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user

$\huge{\underline{\mathtt{\red{Q}\pink{U}\green{E}\blue{S}\purple{T}\orange{I}\red{0}\pink{N}}}}$

$\sf{if \: a \: b \: c \: and \: d \: are \: in \: proportion \: prove \: that} \\ \sf{abcd( \: \frac{1}{a}^{2} +\frac{1}{b}^{2}\frac{1}{c}^{2} \: \frac{1}{d}^{2})= {a}^{2} {b}^{2} {c}^{2}}$
Please no spam. ​

Answer 1

Step-by-step explanation:

2

Given that,

a+b+c=2,a

2 +b

2 +c

2 =1abc=3

Considering a+b+c=2 and squaring both sides

⟹a2 +b 2 +c 2+ 2ab+2bc+2ca=4

⟹2(ab+bc+ca)=4−1

⟹2(a)(b)(c)( a1 + b1 + c1 )=3

⟹2×3×( a1+b1+ c1)=3

⟹ a1 +b1+ c1

=2

Hope its help..

Answer 2

$\Huge{\texttt{{{\color{Magenta}{⛄A}}{\red{N}}{\purple{S}}{\pink{W}}{\blue{E}}{\green{R}}{\red{♡}}{\purple{࿐⛄}}{\color{pink}{:}}}}}$

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

$\\\tt \longrightarrow\dfrac{a}{b} = \dfrac{c}{d} = k(const)$

$\\\tt \longrightarrow \: a= bk \: \: and \: \: c = dk$

$\\\tt \: \: \longrightarrow \: \: abcd \left( \dfrac{1}{a^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{c^{2}}+ \dfrac{1}{d^{2}} \right)$

✠Now put the values –

$\\\tt \: \: \longrightarrow \: \: (bk)b(dk)d \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left[\dfrac{ {d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2} }{b^{2}k^{2}d^{2}} \right]$

$\\\tt \: \: \red{ \longrightarrow \: \:{d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2}}$

$\\\tt \: \: \red{ \longrightarrow\: \:{d}^{2} +{(dk)}^{2} + {b}^{2} + {(bk)}^{2}}$

$\\\tt \: \: \red{ \longrightarrow \: \:{d}^{2} +c^{2} + {b}^{2} +a^{2}}$

$\\\tt \: \: \red{ \longrightarrow \: \:R.H.S}$

✠Hence Proved

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

$\huge\red{\boxed{\orange{\mathcal{{{\fcolorbox{red}{i}{{\red{@Master}}}}}}}}}$