設某直角三角形斜邊長為c，另外兩邊長分別為a和b, 外接圓、內切圓半徑分別為R和r, R = c 2 = a 2 + b 2 2 ≥ 2 a b 2 = 4 A 2 = A {\displaystyle R={\frac {c}{2}}={\frac {\sqrt {a^{2}+b^{2}}}{2}}\geq {\frac {\sqrt {2ab}}{2}}={\frac {\sqrt {4A}}{2}}={\sqrt {A}}} r = 2 A a + b + c ≤ 2 A 2 a b + 2 A = 2 A 2 2 A + 2 A = A 1 + 2 = ( 2 − 1 ) A {\displaystyle r={\frac {2A}{a+b+c}}\leq {\frac {2A}{2{\sqrt {ab}}+2{\sqrt {A}}}}={\frac {2A}{2{\sqrt {2A}}+2{\sqrt {A}}}}={\frac {\sqrt {A}}{1+{\sqrt {2}}}}=({\sqrt {2}}-1){\sqrt {A}}} a=b時等號成立
算幾不等式 a + b ≥ 2 a b {\displaystyle a+b\geq 2{\sqrt {ab}}} 柯西不等式 ( a 2 + b 2 ) ( 1 2 + 1 2 ) ≥ ( a × 1 + b × 1 ) 2 → c 2 × 2 ≥ a 2 + b 2 + 2 a b → 2 c 2 ≥ c 2 + 4 A → c 2 ≥ 4 A → c ≥ 2 A {\displaystyle (a^{2}+b^{2})(1^{2}+1^{2})\geq (a\times 1+b\times 1)^{2}\rightarrow c^{2}\times 2\geq a^{2}+b^{2}+2ab\rightarrow 2c^{2}\geq c^{2}+4A\rightarrow c^{2}\geq 4A\rightarrow c\geq 2{\sqrt {A}}}
