If x, y, w, z >0 and $x^4$+$y^4$+$w^4$+$z^4$ <=4 prove 1/$x^4$+1/$y^4$+1/$w^4$+1/$z^4$>=4 I would appreciate suggestions to solve:
If x, y, w, z > 0 and $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 prove the following:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
From plugging in numbers into Excel, it looks like x, y, w, z must be numbers near 1.
I tried to do a simple case:
If a, b, k > 0 and a + b < k prove the following: 1/a + 1/b > k
a + b < k so 1/(a+b) > 1/k
But 1/a + 1/b > 1/(a+b) so 1/a + 1/b > 1/k which is not what I want to prove.
It seems that I would have to impose a condition for the possible values of k.
 A: Thank you very much for the hints. Here is the solution from the hint of achille hui:
We are told that $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 so ($x^4$+$y^4$ + $w^4$ + $z^4$)/4 <= 1
But AM >= HM (or HM <= AM) so:
4/(1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$) <= ($x^4$+$y^4$ + $w^4$ + $z^4$)/4 <= 1
Which means:
4/(1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$) <= 1
Re-arranging:
4 <= 1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$
In other words:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
A: Remember the Cauchy-Schwarz inequality:
$$(x_1y_1+ \ldots +x_ny_n)^2\leq (x_1^2+ \ldots + x_n^2)(y_1^2+\ldots +y_n^2)$$
In this case we have $n=4$. Set $x_1=x^2, x_2=y^2, x_3=z^2, x_4=w^2$ and $y_1=1/x^2, y_2=1/y^2, y_3=1/z^2, y_4=1/w^2$. Now, applying the inequality we have
$$4^2 \leq (x^4+y^4+z^4+w^4)\left( \frac{1}{x^4}+\frac{1}{y^4} + \frac{1}{z^4}+\frac{1}{w^4} \right).$$
Hence,
$$\frac{16}{x^4+y^4+z^4+w^4} \leq \left( \frac{1}{x^4}+\frac{1}{y^4} + \frac{1}{z^4}+\frac{1}{w^4} \right)$$
Now, by hypotesis, we have $x^4+y^4+z^4+w^4\leq 4$, hence $$\frac{1}{x^4+y^4+z^4+w^4}\geq \frac14.$$
Can you conclude?
