How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]If $a,b$ are positive reals such that $a+b<2$ ,then prove that  $$\displaystyle \frac {1}{1+a^2} + \frac {1}{1+b^2} \leq \frac {2}{1+ab}$$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Mathematical Circles[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Inequality involving AM-GM[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Mathematical Circles[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Assume that the given inequality is true for $a>0 , b>0$ and $a+b<2$ . Then proceed .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]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   $a>0 , b>0$  ) .    [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Come up with the simplest form of inequality  i.e. $(a-b)^2 (1-ab) \geq 0$ .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]$(a-b)^2 \geq 0$ 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 . $$\displaystyle a>0 , b>0 \Rightarrow \sqrt{ab}>0 \Rightarrow( 1+ \sqrt{ab})>0 \\ a>0 , b>0 , a+b <2 \Rightarrow 1 > \frac{a+b}{2} \geq \sqrt {ab} \\ \Rightarrow 1 > \sqrt{ab} \\ \Rightarrow ( 1 - \sqrt{ab}) >0 \\ \Rightarrow (1 - \sqrt{ab}) (1+ \sqrt{ab}) >0 \\ \Rightarrow (1 - ab)>0$$[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="7" _address="0.1.0.7"]