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

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- May 6th 2010, 10:46 AMEinStoneEasy 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 AMslovakiamaths
- May 9th 2010, 04:26 AMPlato
- May 9th 2010, 07:00 AMJester
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 AMUnbeatable0
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 $\displaystyle x^4+y^4 > x^2+y^2$.