Prove the inequality, power.

Question

$\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality: $x^4 + y^4 \ge 2$

How I started

1. $(x+y)^2 = 4$
2. $x^2 + y^2 = 4 - 2xy$
3. $(x^2+y^2)^2 - 2(xy)^2 \ge 2$
4. $(4-2xy)^2 - 2(xy)^2 \ge 2$
5. $16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
6. $2(xy)^2 - 16xy + 14 \ge 0$
7. for $t=xy$
8. $2t^2 - 16t + 14 \ge 0$

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.

Answer 1

$x+y = 2\\ (x+y)^2 = 4\\ (x-y)^2 \ge 0$

add them together

$2x^2 + 2y^2 \ge 4\\ x^2 + y^2 \ge 2$

repeat:

$x^4 + y^4 \ge 2$

Answer 2

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!

Answer 3

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, $$x^4+y^4\geq 4-2=2$$

Answer 4

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