# Easy inequality

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

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

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

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

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

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

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

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

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

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

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

or

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

Completing the square gives

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

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

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

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

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