# 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
5 years, 8 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.

- 5 years, 8 months ago

actually haven't read this theorem.......can you give me a link where it is written in easy and understandable manner....plzzzz

- 5 years, 8 months ago

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...

- 5 years, 8 months ago

thanks :)

- 5 years, 8 months ago

Are u missing any information?

- 5 years, 8 months ago

nup

- 5 years, 8 months ago

ans is c

- 5 years, 8 months ago

how did u got the answer?

- 5 years, 8 months ago