# Problem

• Jul 5th 2006, 04:20 AM
sstr
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.
• Jul 5th 2006, 04:55 AM
Quick
Try proving
$(1-a^2)^2-(1-b^2)^2\geq\frac{a^2+b^2}{2}$
• Jul 5th 2006, 05:05 AM
Soroban
Hello, sstr!

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

Quote:

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

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

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

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

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

Now determine its minimum value.

• Jul 5th 2006, 05:14 AM
CaptainBlack
Quote:

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.

$a^2+b^2=1$,

so:

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

Hence:

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

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

$
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
• Jul 10th 2006, 12:25 PM
ThePerfectHacker
You can convert,
$a^4+b^4$ as,
$a^4+(b^2)^2$
Thus,
$a^4+(1-a^2)^2$
Open,
$a^4+1-2a^2+a^4$
Thus,
$2a^4-2a^2+1$
Now let, $x=a^2$ thus,
$2x^2-2x+1$
And all you need to do is work with a quadradic (you know the max and min point)