MHB Prove $1<sin\,\theta + cos\,\theta \leq \sqrt 2$ for $0<\theta <90^o$

[Albert1]

$0<\theta <90^o$
using geometry to prove the following:
$1<sin\,\, \theta + cos\,\,\theta \leq \sqrt 2$

 

[Albert1]

$0<\theta <90^o$
using geometry to prove the following:
$1<sin\,\, \theta + cos\,\,\theta \leq \sqrt 2$

my soluion

Spoiler

$A,B,C,D$ are four points on a circle, with diameter $\overline {AC}= 1$
let :$\angle DAC=\theta , \angle CAB=\angle BCA=45^o $, then we have:
$\overline {AD}=cos\theta, \overline {CD}=sin\theta, \overline {AB}=\overline {BC}=\dfrac {\sqrt 2}{2}$
it is easy to see $cos\,\theta +sin\, \theta =\overline {AD}+\overline {CD}>\overline {AC}=1$
also by Ptolemy's theorem we have :
$\overline {AB}\times \overline {CD}+\overline {BC}\times\overline {AD}=\overline {AC}\times \overline {BD}=\overline {BD}\leq 1$
that is :$\dfrac {\sqrt 2}{2} sin\, \theta+\dfrac {\sqrt 2}{2} cos\, \theta\leq 1$
or $sin \theta+cos \theta \leq \sqrt 2$