we have by AM GM inaquality
a^2+1 \ge\ 2a \cdots (1)
b^2+1 \ge\ 2b \cdots (2)
c^2+1 \ge\ 2c \cdots (3)
further
\dfrac{a+b+c}{3} \ge \sqrt[3]{ abc}
or \dfrac{a+b+c}{3} \ge 1
or a+b+c \ge 3
or (a+b+c-3 )\ge 0\cdots(4)
adding (1) (2) and (3) we get
a^2+b^2+c^2 + 3 \ge 2(a+b+c)
or a^2+b^2+c^2 \ge a+b+c + (a+b+c-3)
using (4) we get
a^2+b^2+c^2 \ge a+b+c
No comments:
Post a Comment