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
Difficulty Level
Medium
Suggested Book
Mathematical Circles

Start with hints

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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