# Inequality Proof

1. Apr 14, 2007

### siddharthmishra19

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.

Last edited: Apr 14, 2007
2. Apr 14, 2007

### AlephZero

Write sin(y) + cos(y) in the form A sin(y+p).

3. Apr 14, 2007

### siddharthmishra19

Sorry I'm quite weak in trigonometry... could you elaborate?

4. Apr 14, 2007

### HallsofIvy

Staff Emeritus
Csin(y+ B)= Csin(y)cos(B)+ Ccos(y)sin(B) = C cos(B)sin(y)+ C sin(B)cos(y)

You want to find C and B so that C cos(B)= 1 and C sin(B)= 1. Of course, then C2cos2(B)+ C2sin2(B)= C2= 1+ 1= 2. That is, $C= \sqrt{2}$ so that C cos(B)= 1 becomes $\sqrt{2}cos(B)= 1$ and
$$cos(B)= \frac{1}{\sqrt{2}}$$

5. Apr 14, 2007

### siddharthmishra19

Any other simpler methods?

6. Apr 14, 2007