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Try this one, am not getting it.....

If
a^2+b^2+c^2=1
then
ab+bc+ca lies in.....?

a)[1/2,2]
b)[-1,2]
c)[-1/2,1]
d)[-1,1/2]

plzzzz its urgent thankssss in advance....

Note by Riya Gupta
4 years, 3 months ago

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i'm assuming the are all real. Using cauchy-schwarz inequality, we get; \( (a^2+b^2+c^2)(b^2+c^2+a^2) \geq (ab+bc+ca)^2\) hence \(1 \geq (ab+bc+ca)^2 \) therefore, \( 1 \geq ab+bc+ca \geq -1\) Now we could also use \( (a+b+c)^2 \geq 0\) resulting in; \( a^2+b^2+c^2+2(ab+bc+ca) \geq 0\) Hence; \( ab+bc+ca \geq -0.5\) So we get C. Kee Wei Lee · 4 years, 3 months ago

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@Kee Wei Lee actually haven't read this theorem.......can you give me a link where it is written in easy and understandable manner....plzzzz Riya Gupta · 4 years, 3 months ago

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@Riya Gupta Here's a link with some proofs of it:http://rgmia.org/papers/v12e/Cauchy-Schwarzinequality.pdf The second proof is pretty simple and easy... i think... Kee Wei Lee · 4 years, 3 months ago

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@Kee Wei Lee thanks :) Riya Gupta · 4 years, 3 months ago

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ans is c Bhuvnesh Goyal · 4 years, 3 months ago

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@Bhuvnesh Goyal how did u got the answer? Riya Gupta · 4 years, 3 months ago

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Are u missing any information? Aditya Parson · 4 years, 3 months ago

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@Aditya Parson nup Riya Gupta · 4 years, 3 months ago

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