# ~~~hard problem---inequalities~~

• Jul 25th 2006, 12:55 AM
fring
~~~hard problem---inequalities~~
hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

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 1/2

plz help!!! i reeli dun understand the q
thanx guys!!!!! :)
• Jul 25th 2006, 01:53 AM
CaptainBlack
Quote:

Originally Posted by fring
hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

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 1/2

plz help!!! i reeli dun understand the q
thanx guys!!!!! :)

Since $\displaystyle a^2+b^2=1$

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

The minimum of $\displaystyle a^4+(1-a^2)^2$, if it exists, occurs either at an end point of the valid interval for $\displaystyle a$ or at a stationary point of $\displaystyle a^4+(1-a^2)^2$. A quick check shows the minimum does not occur at $\displaystyle a=\pm 1$, so it must occur at a stationary point.

The stationary points of this occur when:

$\displaystyle \frac{d}{da}a^4+(1-a^2)^2=0$

which simplifies to:

$\displaystyle 8a^3-2a=0$

so either $\displaystyle a=0$, or $\displaystyle a=\pm 1/\sqrt{2}$.

Now we need to classifty these stationary points. Examining the second derivative of $\displaystyle a^4+(1-a^2)^2$ at the relevant points shows that $\displaystyle a=\pm 1/\sqrt{2}$ corresponds to a local minimum and that minimum is $\displaystyle 1/2$, and the other point corresponds to a local maximum. Comparison between all the maxima, and minima shows that these are all global maxima and minima, so:

if $\displaystyle a^2+b^2=1$

$\displaystyle a^4+b^4 \ge 1/2$

RonL
• Jul 25th 2006, 09:07 AM
ThePerfectHacker
Quote:

Originally Posted by fring
hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

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 1/2

plz help!!! i reeli dun understand the q
thanx guys!!!!! :)

You have,
$\displaystyle a^2+b^2=1$
Then,
$\displaystyle b^2=1-a^2$
Thus,
$\displaystyle a^4+b^4=a^4+(b^2)^2$
Substitute,
$\displaystyle a^4+(1-a^2)^2$
Open sesame,
$\displaystyle a^4+1-2a^2+a^4=2a^4-2a^2+1$
Let,
$\displaystyle x=a^2$ then,
$\displaystyle 2x^2-2x+1$
This is parabola with minimum point cuz, (2>0) which occurs at,
$\displaystyle x=-\frac{-2}{2\cdot 2}=1/2$
Its minimum value is,
$\displaystyle 2\left( \frac{1}{2} \right)^2-2\left( \frac{1}{2} \right)+1=\frac{1}{2}-1+1=\frac{1}{2}$
Thus, you have that,
$\displaystyle a^4+b^4\geq 1/2$
(Since, the smallest possible value of this expression is 1/2 then any other value [which is more] must be greater than 1/2)