Show that the inequality is true | Geometric Mean

michonamona
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Homework Statement



Let r_{1}, r_{2}, ... , r_{n} be strictly positive numbers. Show that the inequality

(1+R_{G})^{n} \leq V

is true. Where R_{G} = (r_{1}r_{2}...r_{n})^{1/n} and V= \Pi_{k=1}^{n} (1+r_{k})

Homework Equations





The Attempt at a Solution



I've tried taking the log of both sides, as well as expanding out the term. Any insight?

Thanks,
M
 
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Use the arithmetic mean-geometric mean inequality... several times.
 
Any other insights?

The prof hinted that we should use log(1+e^x) and associate r with e^x.
 
That's an entirely different way to approach it. The approach I was thinking of uses the fact that the terms of the right side are the elementary symmetric functions of the r_j.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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