Proof: Clearly for
it is obvious that
. And since
)
is increasing on that interval it stands to reason that
\le\sin\left(x^2\right)\quad 1\le x\le\sqrt{\frac{\pi}{2}})
. Lastly noting that
\le\sin(x)\quad\forall x\in\mathbb{R})
. We may finally deduce that the proposed inequality is definitely true for
/ Therefore let us restrict our attention to
. On this interval
and since
)
is decreasing on this interval we see that
\le\cos\left(x^2\right)\quad 0<x<1)
. We need the following lemma.
Lemma: <x\quad 0<x<1)
Proof: Let
=x-\sin(x))
. Then
=0)
and
=1-\cos(x))
and since
<1\quad 0<x<1)
the conclusion follows.
Using this lemma and the previous deductinos we see that
\le\cos\left(x^2\right)\implies \sin(x)\cos(x)\le\sin(x)\cos\left(x^2\right)\le x\cos\left(x^2\right))
. Now evaluating
\cos(x)dx\le\int_0^z x\cos\left(x^2\right)dx\quad 0<z<1)
gives us
\le\frac{1}{2}\sin\left(z^2\ri ght)\quad 0<z<1)
and the result follows.
LinkBack URL
About LinkBacks. Prove that
\le\sin\left(x^2\right))
it is true that
. So this reduces to proving that
.
Originally Posted by Drexel28 
