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

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Determine the minimal and maximal values the expression $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ can take, where $a,b,c$ are real numbers.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Israel MO 2018, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra, 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="4.0" open="off"]Excursion in Mathematics by Bhaskarcharya Prathisthan[/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"]

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

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Given the expressions, what inequality comes to your mind first? The triangle inequality right? |x| + |y| $\geq$ |x+y|. Can you use this inequality to get a maximum bound?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Maximum Bound: Observe that the maximum bound is got by the triangle inequality as explained. $|a+b| \le |a|+|b|$ $|b+c| \le |b|+|c|$ $|a+c| \le |a|+|c|$ We get, $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$ Never forget to mention the equality case: a = b = c is the equality case.  [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]What about the minimum inequality? The idea is that you can observe that keeping the denominator constant, can you reduce the numerator. Let's take the case of the |a| = |b| = |c| = 1. Now, the expression is maximized when a = b = c = 1 or -1. So, obviously one must be positive or two must be negative or vice-versa. In either case, we get $\frac{2}{3}$. Okay, then maybe we need to deal with the signs and stuff to get a hold on the minimum. Let's fix the signs of a,b,c then, we can break the bonds of the modulus. Let's proceed to the next hint.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Let $a$, $b$, and $c$ be arbitrary real numbers, not all of them equal $0$. By flipping signs, we can assume that at least two of $a$, $b$, and $c$ are non-negative. Actually, without loss of generality, we can assume that $a, b\geq 0$. $3|a + b| + 3|b + c| + 3|c + a| \geq 3a + 3b + 3|b+c| + 3|a+c| \geq 2(a+b) + (a + |a + c|) + (b + |b + c|)$ $= 2(|a| + |b|) + (|-a| + |a + c|) + (|-b| + |b + c|) \geq 2(|a| + |b|) + |c| + |c| = 2|a| + 2|b| + 2|c|$ We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$. The minimum is $\frac{2}{3}$, which is attained for $a = b = 1$ $c = -1$.

# Watch video

[/et_pb_text][et_pb_code _builder_version="3.26.4"]

# Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

# Knowledge Partner  