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.

We need to prove that $\frac{x^4+y^4}{2}\geq\left(\frac{x+y}{2}\right)^4,$
which  enough to prove for non-negatives $x$ and $y$.
Let $x^2+y^2=2uxy$.
Hence, $u\geq1$ and we need to prove that $2u^2-1\geq\left(\frac{u+1}{2}\right)^2,$
which is $(u-1)(7u+5)\geq0$. 
Done!
