Difference between revisions of "Titu's Lemma"
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<cmath> \frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }. </cmath> | <cmath> \frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }. </cmath> | ||
− | It is a direct consequence of Cauchy-Schwarz | + | It is a direct consequence of [[Cauchy-Schwarz inequality]]. |
+ | |||
+ | Equality holds when <math>a_i = kb_i</math> for <math>1 \leq i \leq n</math>. | ||
Titu's lemma is named after Titu Andreescu and is also known as T2 lemma, Engel's form, or Sedrakyan's inequality. | Titu's lemma is named after Titu Andreescu and is also known as T2 lemma, Engel's form, or Sedrakyan's inequality. | ||
+ | |||
+ | ==Examples== | ||
+ | ===Example 1=== | ||
+ | Given that positive reals <math>a</math>, <math>b</math>, and <math>c</math> are subject to <math>a + b + c = 1</math>, find the minimum value of <math>\frac{1}{a} + \frac{4}{b} + \frac{9}{c}</math>. (Source: [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]) | ||
+ | |||
+ | ====Solution==== | ||
+ | This is a somewhat standard application of Titu's lemma. Notice that <cmath>\frac{1}{a} + \frac{4}{b} + \frac{9}{c} = \frac{1^2}{a} + \frac{2^2}{b} + \frac{3^2}{c}.</cmath> When solving problems with Titu's lemma, the goal is to get perfect squares in the numerator. Now, we can apply the lemma. <cmath>\frac{1^2}{a} + \frac{2^2}{b} + \frac{3^2}{c}</cmath> <cmath>\geq \frac{(1+2+3)^2}{a + b + c}</cmath> <cmath>= \boxed{36}</cmath> | ||
+ | |||
+ | ===Example 2=== | ||
+ | Prove [https://artofproblemsolving.com/wiki/index.php/Nesbitt%27s_Inequality Nesbitt's Inequality]. | ||
+ | |||
+ | ====Solution==== | ||
+ | For reference, Nesbitt's Inequality states that for positive reals <math>a</math>, <math>b</math>, and <math>c</math>, <cmath>\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \geq \frac{3}{2}.</cmath> We rewrite as follows. <cmath>\frac{a^2}{ab + ac} + \frac{b^2}{ab + bc} + \frac{c^2}{ac + bc}</cmath> <cmath>\geq \frac{(a + b + c)^2}{2(ab + ac + bc)}</cmath> This is the application of Titu's lemma. <cmath>\geq \frac{3(ab + bc + ac)}{2(ab + ac + bc)}</cmath> This step follows from <math>a^2 + b^2 + c^2 \geq ab + ac + bc \implies (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) \geq 3(ab + bc + ac)</math>. <cmath>= \frac{3}{2}</cmath> <cmath>\blacksquare</cmath> | ||
+ | |||
+ | ===Example 3=== | ||
+ | Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</math>, <math>b_1</math>, <math>b_2</math>, <math>\cdots</math>, <math>b_n</math> be positive real numbers such that <math>a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n</math>. Show that | ||
+ | <cmath> \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} </cmath> ([https://artofproblemsolving.com/wiki/index.php/1991_APMO_Problems/Problem_3 Source]) | ||
+ | |||
+ | ====Solution==== | ||
+ | By Titu's Lemma, | ||
+ | <cmath>\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1 + a_2 + \ldots + a_n)^2}{a_1 + a_2 + ... + a_n + b_1 + b_2 + ... + b_n}</cmath> <cmath>= \frac{(a_1 + a_2 + \ldots + a_n)^2}{2(a_1 + a_2 + ... + a_n)}</cmath> This is valid because <math>a_1 + a_2 + ... + a_n = b_1 + b_2 + ... + b_n</math> (from the problem statement). <cmath>= \frac{a_1 + a_2 + \ldots + a_n}{2}</cmath> <cmath>\blacksquare</cmath> | ||
+ | |||
+ | ==Problems== | ||
+ | ===Introductory=== | ||
+ | *There exists a smallest possible integer <math>N</math> such that <cmath>\frac{(x_1 + 2x_2 + \ldots + 2014x_{2014})^2}{x_1^2 + x_2^2 + \ldots + x_{2014}^2} \leq N</cmath> for all real sequences <math>x_1, x_2, ..., x_{2014}</math>. Find the sum of the digits of <math>N</math>. ([https://brilliant.org/wiki/titus-lemma/ Source]) | ||
+ | |||
+ | ===Intermediate=== | ||
+ | *Let positive real numbers <math>x</math>, <math>y</math>, and <math>z</math> be roots of the polynomial <math>p(x) = x^3 + bx^2 + cx + 2024</math> for some fixed integer <math>b</math>. For this <math>b</math>, there exists an integer <math>k</math> such that, as <math>c</math> varies through the reals, <math>\sqrt{\frac{x^2yz + 3x^2y + 2x^2z + 6x^2 + xy^2z + 3xy^2 + y^2z + 3y^2 + xyz^2 + 2xz^2 + yz^2 + 2z^2}{xyz + 3xy + 2xz + 6x + yz + 3y + 2z + 6}} \geq k</math>. What is the sum of all possible values of <math>p(1)+p(-1)</math>? (Source: [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]) | ||
+ | |||
+ | *Prove that, for all positive real numbers <math>a, b, c,</math> <cmath>\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}.</cmath> ([https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_5 Source]) | ||
+ | |||
+ | ===Olympiad=== | ||
+ | *Let <math>a, b, c</math> be positive real numbers such that <math>abc = 1</math>. Prove that <cmath>\frac{ab}{a^5 + b^5 + ab} + \frac{bc}{b^5 + c^5 + bc} + \frac{ac}{a^5 + c^5 + ac} \leq 1.</cmath> ([https://artofproblemsolving.com/downloads/printable_post_collections/3947 Source]) | ||
+ | |||
+ | *Let <math>a, b, c</math> be positive real numbers such that <math>abc = 1</math>. Prove that | ||
+ | <cmath> \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. </cmath> ([https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_2 Source]) |
Latest revision as of 20:10, 29 April 2024
Titu's lemma states that:
It is a direct consequence of Cauchy-Schwarz inequality.
Equality holds when for
.
Titu's lemma is named after Titu Andreescu and is also known as T2 lemma, Engel's form, or Sedrakyan's inequality.
Contents
Examples
Example 1
Given that positive reals ,
, and
are subject to
, find the minimum value of
. (Source: cxsmi)
Solution
This is a somewhat standard application of Titu's lemma. Notice that When solving problems with Titu's lemma, the goal is to get perfect squares in the numerator. Now, we can apply the lemma.
Example 2
Prove Nesbitt's Inequality.
Solution
For reference, Nesbitt's Inequality states that for positive reals ,
, and
,
We rewrite as follows.
This is the application of Titu's lemma.
This step follows from
.
Example 3
Let ,
,
,
,
,
,
,
be positive real numbers such that
. Show that
(Source)
Solution
By Titu's Lemma,
This is valid because
(from the problem statement).
Problems
Introductory
- There exists a smallest possible integer
such that
for all real sequences
. Find the sum of the digits of
. (Source)
Intermediate
- Let positive real numbers
,
, and
be roots of the polynomial
for some fixed integer
. For this
, there exists an integer
such that, as
varies through the reals,
. What is the sum of all possible values of
? (Source: cxsmi)
- Prove that, for all positive real numbers
(Source)
Olympiad
- Let
be positive real numbers such that
. Prove that
(Source)
- Let
be positive real numbers such that
. Prove that
(Source)