(adsbygoogle = window.adsbygoogle || []).push({}); 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Inequality Proof

**Physics Forums | Science Articles, Homework Help, Discussion**