Question

Suppose that \(f(x) \ : [0,a] \rightarrow R^+\ \) is a function...prove that
\[\int\limits_{0}^{a} f(x) dx \times \int\limits_{0}^{a} \frac{1}{f(x)} dx \ge a^2\]

Answer 1

*

Answer 2

don't use theorems...

Answer 3

is that really a cross product?

Answer 4

no

Answer 5

......I'm thinking Cauchy inequality. Maybe.

Answer 6

Oops sorry can't use theorems.

Answer 7

if you multiply them togethere because they both have the same endpoints we obtain

the integral of [1] from 0 to a >= a^2 now we have

a - 0>=a^2 (right side by the fundamental theorem of calc) which is a>=a^2 hmm strange

Answer 8

\[\frac{f(y)}{f(x)}+\frac{f(x)}{f(y)} \ge 2(z+1/z >=2 \text { if } z>0)\]
then \[\int\limits_{0}^{y}\frac{f(y)}{f(x)}dx+\int\limits_{0}^{y}\frac{f(x)}{f(y)}dx \ge \int\limits_{0}^{y}2dx=2y\text{**}\]
\[f(y)\int\limits_{0}^{y}\frac{1}{f(x)}dx+\frac{1}{f(y)}\int\limits_{0}^{y}f(x)dx\]
=\[\frac{d(\int\limits_{0}^{y}f(x)dx.\int\limits_{0}^{y} \frac{1}{f(x)}dx=F(y))}{dy} \]

we will integrate (**) from 0 to a
then \[F(a)-F(0) \ge \int\limits_{0}^{a}2ydy=a^2 \text{ with } F(0)=0\]
then \[F(a)=\int\limits_{0}^{a}f(x)dx.\int\limits_{0}^{a}\frac{1}{f(x)}dx \ge a^2\]
notes : y in [0;a]
and f > 0
I didn't make it clear enough !because I have a terrible headache :( !

Answer 9

\[I=\int\limits_{0}^{a} f(x) dx \times \int\limits_{0}^{a} \frac{1}{f(x)} dx=\int\limits_{0}^{a} f(x) dx \times \int\limits_{0}^{a} \frac{1}{f(y)} dy=\int\limits_{0}^{a} \int\limits_{0}^{a} \frac{f(x)}{f(y)} dxdy\]
and

\[2I =\int\limits_{0}^{a} \int\limits_{0}^{a} \frac{f(x)}{f(y)} dxdy+\int\limits_{0}^{a} \int\limits_{0}^{a} \frac{f(y)}{f(x)} dxdy\\=\int\limits_{0}^{a} \int\limits_{0}^{a} \left( \frac{f(x)}{f(y)}+\frac{f(y)}{f(x)} \right) dxdy \ge \int\limits_{0}^{a} \int\limits_{0}^{a} 2 \ dxdy=2a^2 \ \ \ \ so \ \ I \ge a^2\]

Answer 10

@mukushla One should put some conditions on f
like f continuous
f is never zero.

Answer 11

i checked my notes...u r right sir...i forgot to mention that conditions...:)