#### FriendlyAsian

##### New member

- Joined
- Mar 22, 2019

- Messages
- 9

\(\displaystyle \dfrac{1}{ab} +\dfrac{1}{(a^2+b^2)} \geq 6 \)

- Thread starter FriendlyAsian
- Start date

- Joined
- Mar 22, 2019

- Messages
- 9

\(\displaystyle \dfrac{1}{ab} +\dfrac{1}{(a^2+b^2)} \geq 6 \)

- Joined
- Nov 12, 2017

- Messages
- 3,313

- Joined
- Jan 29, 2005

- Messages
- 7,948

If \(\displaystyle \{a,b\}\subseteq\mathbb{Z}^+\) then \(\displaystyle a+b\ne 1\).

\(\displaystyle \dfrac{1}{ab} +\dfrac{1}{(a^2+b^2)} \geq 6 \)

Please correct your post.

- Joined
- Mar 22, 2019

- Messages
- 9

Besides I've figured out a solution

We have \(\displaystyle

4ab\leq (a+b)^{2}=1. \Rightarrow \dfrac{1}{ab}\geq 4.

\).

=>Left side=\(\displaystyle =\dfrac{1}{ab} +\dfrac{1}{a^2+b^2} = (\dfrac{1}{2ab}+\dfrac{1}{a^{2}+b^{2}})+\dfrac{1}{2ab}\geq \dfrac{4}{(a+b)^{2}}+\dfrac{1}{2}.4=4+2=6. \)

- Joined
- Nov 12, 2017

- Messages
- 3,313

Conditions are important! Time has been wasted by getting it wrong.