1. Problem

Let a and b be positive real numbers such that

a^2 + b^2 = 1

Prove that
a^4 + b^4 is greater than or equal to 0.5

Would someone be so kind as to point me in the right direction?

P.S. Don't give me the answer just yet. I want to see if I can do it alone. Just give me an idea of how I should go about doing this.

2. Try proving
$\displaystyle (1-a^2)^2-(1-b^2)^2\geq\frac{a^2+b^2}{2}$

3. Hello, sstr!

I found a method . . . I'll give you the game plan.

Let $\displaystyle a$ and $\displaystyle b$ be positive real numbers such that: .$\displaystyle a^2 + b^2 \,= \,1$

Prove that: $\displaystyle a^4 + b^4 \,\geq \,\frac{1}{2}$

We have: .$\displaystyle b^2 \,= \,1 - a^2$

Substitute into .$\displaystyle a^4\,+\,b^4$

We will have a function of $\displaystyle a:\;\;f(a) \;= \;a^2 + (1 - a^2)^2$

Now determine its minimum value.

4. Originally Posted by sstr
Let a and b be positive real numbers such that

a^2 + b^2 = 1

Prove that
a^4 + b^4 is greater than or equal to 0.5

Would someone be so kind as to point me in the right direction?

P.S. Don't give me the answer just yet. I want to see if I can do it alone. Just give me an idea of how I should go about doing this.
$\displaystyle a^2+b^2=1$,

so:

$\displaystyle (a^2+b^2)^2=a^4+b^4+2a^2b^2=1$

Hence:

$\displaystyle a^4+b^4=1-2a^2b^2$

subject to $\displaystyle a^2+b^2=1$ so:

$\displaystyle a^4+b^4=1-2a^2(1-a^2)$.

Now find and classify the extrema of the Right Hand Side of this last equation.

RonL

5. You can convert,
$\displaystyle a^4+b^4$ as,
$\displaystyle a^4+(b^2)^2$
Thus,
$\displaystyle a^4+(1-a^2)^2$
Open,
$\displaystyle a^4+1-2a^2+a^4$
Thus,
$\displaystyle 2a^4-2a^2+1$
Now let, $\displaystyle x=a^2$ thus,
$\displaystyle 2x^2-2x+1$
And all you need to do is work with a quadradic (you know the max and min point)