Nice Inequality

Rijul saini

If $x,y,z$ are positive reals such that $xyz=1$ and
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \ge x+y+z$
Prove that
$\frac{1}{x^k}+\frac{1}{y^k}+\frac{1}{z^k} \ge x^k+y^k+z^k$
For any positive integer $k$

**Posted:** Fri Nov 06, 2009 8:32 am

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peine · Riemann Hypothesis Offline Joined: 07 Aug 2008 Posts: 280

nice but easy,
we have $\frac {1}{x} + \frac {1}{y} + \frac {1}{z}\geq x + y + z$
$\Leftrightarrow xy + yz + zx\geq x + y + z$
$\Leftrightarrow (1 - x)(1 - y)(1 - z)\geq 0$
using the formula $a^k - 1 = (a - 1)(a^{k - 1} + ... + 1)$ we get,
$(1 - x^k)(1 - y^k)(1 - z^k) = (1 - x)(1 - y)(1 - z)(x^{k - 1} + ... + 1)(y^{k - 1} + ... + 1)(z^{k - 1} + ... + 1)\geq 0$
$\Leftrightarrow \frac {1}{x^k} + \frac {1}{y^k} + \frac {z^k} \geq x^k + y^k + z^k$
equality holds when $x = 1$ or $y = 1$ or $z = 1$
using my solution we can also prove that, if $x,y,z > 0$ and $xyz = 1$ and $\frac {1}{x} + \frac {1}{y} + \frac {1}{z}\leq x + y + z$
then $\frac {1}{x^k} + \frac {1}{y^k} + \frac{1}{z^k} \leq x^k + y^k + z^k$ holds for all integer $k$