# $$(ax+by)^2 \leq ax^2+by^2$$ for every a,b which satisfy

1. Jul 8, 2011

### estro

$$(ax+by)^2 \leq ax^2+by^2$$ for every a,b which satisfy $$0\leq a,b\leq1$$ and a+b=1.

My book consider this as trivial however I have hard time to prove this, will appreciate your help.

2. Jul 8, 2011

### Mentallic

Re: inequality

Expand the left side, let b=1-a and simplify. See if you can convert all factors involving a into a2-a

3. Jul 8, 2011

### estro

Re: inequality

I have tried many things including this one, how should I approach these kind of problems?
What is the intuition? Is the solution process of these kind of problems really involves trial and error method or I miss something?

4. Jul 8, 2011

### estro

Re: inequality

Eventually i came to this nasty prove:

$$ax^2+by^2-(ax+by)^2=ax^2+by^2-a^2x^2-2abxy-b^2y^2=(a-a^2)x^2-2abxy+(b-b^2)y^2=(a-a^2)x^2-2(a-a^2)xy+(b-b^2)y^2=(a-a^2)[(x-y)^2-y^2]+(b-b^2)y^2= (a-a^2)(x-y)^2+y^2[b-b^2+a^2-a]=(a-a^2)(x-y)^2+y^2[2ab] \geq 0$$

Is there a more efficient approaches to such problems?

5. Jul 9, 2011

### Mentallic

Re: inequality

Yes, use the fact that a+b=1, thus b=1-a. Use this conversion early on.