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**1. The Problem**

For all positive real values of a,b and c such that a²+b²+c²=1, prove that

a(b+c) < [1/sqrt(2)]

**2. My Attempt...**

There are probably numerous ways to solve this, but i tried this way...

let a = sin(x)

b = cos(x)sin(y)

c = cos(x)cos(y)

a²+b²+c² = sin²(x)+cos²(x)[sin²(y)+cos²(y)] = 1

This satisfies the above

so a(b+c) = sin(x)cos(x)[sin(y)+cos(y)]

=(1/2)sin2(x)[sin(y)+cos(y)]...

I am sure you can somehow prove that the max value of the give is less than than the value given in the question i.e. 1/sqrt(2)...

I would appreciate any help... with any other methods.

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