a+b+c>=1/a+1/b+1/c

For discussing Olympiad Level Algebra (and Inequality) problems
Katy729
Posts: 40
Joined: Sat May 06, 2017 2:30 am

a+b+c>=1/a+1/b+1/c

Unread post by Katy729 » Sat Jul 01, 2017 3:31 pm

Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$

Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]

Katy729
Posts: 40
Joined: Sat May 06, 2017 2:30 am

Re: a+b+c>=1/a+1/b+1/c

Unread post by Katy729 » Wed Sep 06, 2017 6:02 pm

Someone? :(

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