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....

#HelpMe! #Advice #MathProblem

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Comments

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 - 8 years, 1 month ago

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

A Former Brilliant Member - 8 years, 1 month 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...

Kee Wei Lee - 8 years, 1 month ago

@Kee Wei Lee – thanks :)

A Former Brilliant Member - 8 years, 1 month ago

Are u missing any information?

Aditya Parson - 8 years, 1 month ago

nup

A Former Brilliant Member - 8 years, 1 month ago

ans is c

bhuvnesh goyal - 8 years, 1 month ago

how did u got the answer?

A Former Brilliant Member - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$