# Easy inequality

• May 6th 2010, 10:46 AM
EinStone
Easy inequality
Let $x,y$ be real. Show that if $x^2 + y^2 > 2$, then $x^4 + y^4 > x^2 + y^2$
• May 9th 2010, 04:12 AM
slovakiamaths
Quote:

Originally Posted by EinStone
Let $x,y$ be real. Show that if $x^2 + y^2 > 2$, then $x^4 + y^4 > x^2 + y^2$

As $x^2 + y^2 > 2$ it means x and y are both greater than 1. so,
$x^4>x^2 ...............(i)$
$y^4>y^2................(ii)$
adding (i) & (ii), we have
$x^4 + y^4 > x^2 + y^2$
• May 9th 2010, 04:26 AM
Plato
Quote:

Originally Posted by slovakiamaths
As $\color{red}x^2 + y^2 > 2$ it means x and y are both greater than 1. so,

What makes you think that is true?
It is not true: $x=2~\&~y=0.5$
• May 9th 2010, 07:00 AM
Jester
Quote:

Originally Posted by EinStone
Let $x,y$ be real. Show that if $x^2 + y^2 > 2$, then $x^4 + y^4 > x^2 + y^2$

If $x = r \cos \theta$ and $y = r \sin \theta$, then you would like to establish that

$r^2 \left( \cos^4 \theta + \sin^4 \theta\right) > 1 \; \text{if}\; r^2 > 2$ or

$
\cos^4 \theta + \sin^4 \theta \ge \frac{1}{2}.
$

Since $\cos^2\theta = 1 - \sin^2 \theta$ then we wish to prove that

$
\sin^4 \theta +(1 - \sin^2 \theta)^2 \ge \frac{1}{2},
$

or

$
2 \sin^4 \theta - 2\sin^2 \theta + 1 \ge \frac{1}{2}
$

Completing the square gives

$
\left( \sin^2 \theta - \frac{1}{2}\right)^2 + \frac{1}{4} \ge \frac{1}{4}
$

which obviously is true.
• May 9th 2010, 08:45 AM
Unbeatable0
Assume, without loss of generality, that $x^2\ge 1 \ge y^2$.

$x^2+y^2 > 2 \Leftrightarrow x^2-1 > 1-y^2$,

where $x^2-1 \ge 0$ and $1-y^2 \le 0$,

so $x^2(x^2-1)>y^2(1-y^2)$, and thus $x^4+y^4 > x^2+y^2$.