Regional Math Olympiad, 2019 Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Inequality[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="off"]Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
The clue: Number 5 in the right-hand side!We will be applying AM-GM inequality. But first, to get the 5 on the right, we need 5 terms in the left (or bunch of five terms).Â
Try to use a+b+c =1 to cook it up!
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]In the first term's denominator, we have \( a^2 + b^3 + c^3 \).Multiply 1 to \( a^2 \) that is multiply by a + b + c (nothing changes because, multiplying by does not change anything).Hence we have \(a^2 \cdot 1 + b^3 + c^3 =  a^2 ( a + b + c) + b^3 + c^3 \)Expanding we have \( a^3 + b^3 + c^3 + a^2 b + a^2 c \)Now apply AM - GM inequality to this we have $$ \frac{ a^3 + b^3 + c^3 + a^2 b + a^2 c}{5} \geq (a^3 \cdot b^3 \cdot c^3 \cdot a^2 b \cdot a^2 c )^{1/5} $$Therefore we have $$ a^3 + b^3 + c^3 + a^2 b + a^2 c \geq 5 \cdot (a^7 b^4 c^4)^{1/5} $$Taking the reciprocal we have and noting that the left hand side is still \( a^2 + b^3 + c^3 \) we have $$ \frac {1} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{7/5} b^{4/5} c^{4/5} } $$Multiplying the numerator and denominator by a we have the desired expression in the left.$$ \frac {a} {a^2 + b^3 + c^3} \leq \frac {a}{5 \cdot (a^{7/5} b^{4/5} c^{4/5} } $$Simplifying$$ \frac {a} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } $$Now try computing the same for the other two terms on the left.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]$$ \frac {a} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } $$$$ \frac {b} {b^2 + c^3 + a^3} \leq \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } $$$$ \frac {c} {c^2 + a^3 + b^3} \leq \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} } $$Adding we have $$ \frac {a} {a^2 + b^3 + c^3} + \frac {b} {b^2 + c^3 + a^3} + \frac {c} {c^2 + a^3 + b^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } +  \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} } $$Now apply AM- GM Inequality one more time to the left hand term. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]$$ \frac{ \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } +  \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }}{3} \leq ( \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } \cdot \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } \cdot \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} })^{1/3} $$Simplifying we have $$ \frac{ \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } +  \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }}{3} \leq ( \frac {1}{5^3 \cdot (a^{10/5} b^{10/5} c^{10/5} } )^{1/3} $$Simplifying further we have$$ \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } +  \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} } \leq 3 \cdot \frac {(abc)^{1/3}}{5abc} $$Finally we know that \( 3 \cdot (abc)^{1/3} \leq 1 \). Why? Apply AM-GM to a, b, c$$ \frac{a+b+c}{3} \geq (abc)^{1/3} $$Since a + b + c = 1 we have the result.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.in/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
