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.

Also we can use C-S twice:
$x^4+y^4=\frac{1}{2}(1+1)(x^4+y^4)\geq\frac{1}{2}(x^2+y^2)^2=\frac{1}{8}((1+1)(x^2+y^2))^2\geq\frac{1}{8}\left((x+y)^2\right)^2=2$
