R ≥ ( 1 + 2 ) r {\displaystyle R\ge (1+\sqrt{2})r}
Kako je
R = c 2 {\displaystyle R=\frac{c}{2}}
r = a + − c 2 {\displaystyle r=\frac{a+-c}{2}}
Emmerichovu nejednakost
svodimo na
c 2 ≥ ( 1 + 2 ) r = a + − c 2 {\displaystyle \frac{c}{2} \ge (1+\sqrt{2})r=\frac{a+-c}{2}}
c ≥ ( 1 + 2 ) ( a + b − c ) / ( 2 − 1 ) {\displaystyle c\ge(1+\sqrt{2})(a+b-c) / (\sqrt{2}-1)}
c ( 2 − 1 ) ≥ ( a + b − c ) {\displaystyle c(\sqrt{2}-1)\ge (a+b-c )}
a + b ≤ c 2 {\displaystyle a+b\le c\sqrt{2}}
Koristeci geometrijsku i kvadratnu nejednakost imamo
a + b 2 ≤ a 2 + b 2 2 {\displaystyle \frac{a+b}{2}\le \sqrt{\frac{a^2+b^2}{2}}}
a + b ≤ 2 ( a 2 + b 2 ) = 2 c 2 ) = c 2 {\displaystyle a+b \le\sqrt{2(a^2+b^2)}=\sqrt{2c^2)}= c\sqrt{2}}
Ovo znaci da je Emmerichova nejednakost tacna. Jednakost vazi za a = b {\displaystyle a = b} tj za pravougli jednakokraki trougao.