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$\{ x,y \in\Bbb R\ \}$ 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.

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  • $\begingroup$ Do you know the power means inequality? $\endgroup$ – Bernard Mar 10 '17 at 23:12
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$x+y = 2\\ (x+y)^2 = 4\\ (x-y)^2 \ge 0$

add them together

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

repeat:

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

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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!

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  • $\begingroup$ Why is it enough to prove for non-negative values of x and y? $\endgroup$ – mrnovice Mar 10 '17 at 22:56
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    $\begingroup$ @mrnovice If $xy>0$ and $x<0$, $y<0$ so after replacing $x\rightarrow-x$ and $y\rightarrow-y$ our inequality does not changed. If $xy<0$ and for example $x>0$, $y<0$ and $x+y\geq0$ so after replacing $y\rightarrow-y$ we need to prove that $\frac{x^4+y^4}{2}\geq\left(\frac{x-y}{2}\right)^4$ for positives $x$ and $y$ and since $\left(\frac{x-y}{2}\right)^4<\left(\frac{x+y}{2}\right)^4$, it's enough to prove $\frac{x^4+y^4}{2}\geq\left(\frac{x+y}{2}\right)^4$ $\endgroup$ – Michael Rozenberg Mar 10 '17 at 23:03
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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, $$x^4+y^4\geq 4-2=2$$

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

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