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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, 5 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, 5 months ago

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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, 5 months ago

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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, 5 months ago

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@Kee Wei Lee thanks :)

Riya Gupta - 4 years, 5 months ago

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ans is c

Bhuvnesh Goyal - 4 years, 5 months ago

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how did u got the answer?

Riya Gupta - 4 years, 5 months ago

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Are u missing any information?

Aditya Parson - 4 years, 5 months ago

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nup

Riya Gupta - 4 years, 5 months ago

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