ineq (1 Viewer)

shuning · Jul 18, 2009

Given that x + y = p prove if x>0, y>0

$\frac{1}{x}+\frac{1}{y}\geq \frac{4}{p}$
and

$\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{8}{p^2}$

lolokay · Jul 18, 2009

just work backwards and use the (x-y)² >= 0 inequality

$(x-y)^2\geq 0\\ (x+y)^2\geq 4xy\\ \frac{y+x}{xy}\geq \frac{4}{x+y}\\ \frac{1}{x}+\frac{1}{y}\geq \frac{4}{p}$

and

$(x-y)^2\geq 0\\ x^2+y^2\geq 2xy\\ and\ \\ (x+y)^2\geq 4xy\\ (x^2+y^2)(x+y)^2\geq 8x^2y^2\\ \frac{x^2+y^2}{x^2y^2}\geq \frac{8}{(x+y)^2}\\ \frac{1}{x^2}+\frac{1}{y^2}\geq \frac{8}{p^2}$

ineq (1 Viewer)

shuning

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lolokay

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