# Understand the problem

If $$a,b$$ are positive reals such that $$a+b<2$$ ,then prove that $$\displaystyle \frac {1}{1+a^2} + \frac {1}{1+b^2} \leq \frac {2}{1+ab}$$
##### Source of the problem
Mathematical Circles
##### Topic
Inequality involving AM-GM
Medium
##### Suggested Book
Mathematical Circles

Do you really need a hint? Try it first!

Assume that the given inequality is true for $$a>0 , b>0$$ and $$a+b<2$$ . Then proceed .
In order to simplyfy the given inequality multiply both the sides by $$(1+a^2)(1+b^2)(1+ab)$$ (as its a positive quantity and it is directly coming from $$a>0 , b>0$$ ) .
Come up with the simplest form of inequality i.e. $$(a-b)^2 (1-ab) \geq 0$$ .
$$(a-b)^2 \geq 0$$ as $$a,b \in{R}$$ . And to get $$(1-ab)>0$$ use the well known inequality for positive reals i.e. $$AM \geq GM$$ and the still unused inequality i.e $$a+b <2$$ also . $$\displaystyle a>0 , b>0 \Rightarrow \sqrt{ab}>0 \Rightarrow( 1+ \sqrt{ab})>0 \\ a>0 , b>0 , a+b <2 \Rightarrow 1 > \frac{a+b}{2} \geq \sqrt {ab} \\ \Rightarrow 1 > \sqrt{ab} \\ \Rightarrow ( 1 – \sqrt{ab}) >0 \\ \Rightarrow (1 – \sqrt{ab}) (1+ \sqrt{ab}) >0 \\ \Rightarrow (1 – ab)>0$$

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