Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started


*

*$(x+y)^2 = 4$

*$x^2 + y^2 = 4 - 2xy$

*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$

*$(4-2xy)^2 - 2(xy)^2 \ge 2$

*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$

*$2(xy)^2 - 16xy + 14 \ge 0$

*for $t=xy$

*$2t^2 - 16t + 14 \ge 0$


It isn't always true, I think that I should have a assumption for $t$, but I don't know how should I do this.

Using Holder's inequality we get that 
$$(1+1+1+1)^{3/4}(x^4+y^4+1+1)^{1/4} \geq |x|+|y| +1+1 \geq4  $$
Therefore,
$$(x^4+y^4+1+1)^{1/4}\geq 4^{1/4}$$
We conclude,
$$x^4+y^4\geq 4-2=2$$
