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Thursday, October 9, 2014

2014/089) show that if abc = 1 then a^2+b^2+c^2\ge a + b + c

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

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