Có a²+ b²+ c²=1
Có (a+b+c)² ≤ 3(a²+b²+c²)= 3
=> a+b+c ≤ $\sqrt[]{3}$
Có A= $\sqrt[]{a+b²}$ + $\sqrt[]{b+c²}$+ $\sqrt[]{c+a²}$
Áp dụng bđt bunhia
A² ≤ (1+1+1)(a+b²+ b+ c²+ c+ a²)
≤ 3( a+b+c+1)
≤ 3($\sqrt[]{3}$ +1)= 3$\sqrt[]{3}$ +3
=> A ≤ $\sqrt[]{3\sqrt[]{3} +3}$
Dấu "=" xảy ra <=> a=b=c = $\frac{1}{\sqrt[]{3}}$