Difference between revisions of "Titu's Lemma"

Latest revision as of 19:10, 29 April 2024

Titu's lemma states that:

$\frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }.$

It is a direct consequence of Cauchy-Schwarz inequality.

Equality holds when $a_i = kb_i$ for $1 \leq i \leq n$ .

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 $a$ , $b$ , and $c$ are subject to $a + b + c = 1$ , find the minimum value of $\frac{1}{a} + \frac{4}{b} + \frac{9}{c}$ . (Source: cxsmi)

Solution

This is a somewhat standard application of Titu's lemma. Notice that $\frac{1}{a} + \frac{4}{b} + \frac{9}{c} = \frac{1^2}{a} + \frac{2^2}{b} + \frac{3^2}{c}.$ When solving problems with Titu's lemma, the goal is to get perfect squares in the numerator. Now, we can apply the lemma. $\frac{1^2}{a} + \frac{2^2}{b} + \frac{3^2}{c}$ $\geq \frac{(1+2+3)^2}{a + b + c}$ $= \boxed{36}$

Example 2

Prove Nesbitt's Inequality.

Solution

For reference, Nesbitt's Inequality states that for positive reals $a$ , $b$ , and $c$ , $\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \geq \frac{3}{2}.$ We rewrite as follows. $\frac{a^2}{ab + ac} + \frac{b^2}{ab + bc} + \frac{c^2}{ac + bc}$ $\geq \frac{(a + b + c)^2}{2(ab + ac + bc)}$ This is the application of Titu's lemma. $\geq \frac{3(ab + bc + ac)}{2(ab + ac + bc)}$ This step follows from $a^2 + b^2 + c^2 \geq ab + ac + bc \implies (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) \geq 3(ab + bc + ac)$ . $= \frac{3}{2}$ $\blacksquare$

Example 3

Let $a_1$ , $a_2$ , $\cdots$ , $a_n$ , $b_1$ , $b_2$ , $\cdots$ , $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$ . Show that $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}$ (Source)

Solution

By Titu's Lemma, $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1 + a_2 + \ldots + a_n)^2}{a_1 + a_2 + ... + a_n + b_1 + b_2 + ... + b_n}$ $= \frac{(a_1 + a_2 + \ldots + a_n)^2}{2(a_1 + a_2 + ... + a_n)}$ This is valid because $a_1 + a_2 + ... + a_n = b_1 + b_2 + ... + b_n$ (from the problem statement). $= \frac{a_1 + a_2 + \ldots + a_n}{2}$ $\blacksquare$

Problems

Introductory

There exists a smallest possible integer $N$ such that $\frac{(x_1 + 2x_2 + \ldots + 2014x_{2014})^2}{x_1^2 + x_2^2 + \ldots + x_{2014}^2} \leq N$ for all real sequences $x_1, x_2, ..., x_{2014}$ . Find the sum of the digits of $N$ . (Source)

Intermediate

Let positive real numbers $x$ , $y$ , and $z$ be roots of the polynomial $p(x) = x^3 + bx^2 + cx + 2024$ for some fixed integer $b$ . For this $b$ , there exists an integer $k$ such that, as $c$ varies through the reals, $\sqrt{\frac{x^2yz + 3x^2y + 2x^2z + 6x^2 + xy^2z + 3xy^2 + y^2z + 3y^2 + xyz^2 + 2xz^2 + yz^2 + 2z^2}{xyz + 3xy + 2xz + 6x + yz + 3y + 2z + 6}} \geq k$ . What is the sum of all possible values of $p(1)+p(-1)$ ? (Source: cxsmi)

Prove that, for all positive real numbers $a, b, c,$ $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}.$ (Source)

Olympiad

Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $\frac{ab}{a^5 + b^5 + ab} + \frac{bc}{b^5 + c^5 + bc} + \frac{ac}{a^5 + c^5 + ac} \leq 1.$ (Source)

Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.$ (Source)

@@ Line 35: / Line 35: @@
 ===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 + d</math> for some real <math>b</math>, <math>c</math>, and <math>d</math>. There exists some minimum integer <math>k>1</math> such that the equation <math>h = \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}}</math> has some solution in <math>x</math>, <math>y</math>, and <math>z</math> when <math>h=k</math> and no solutions when <math>h<k</math>. This is the case if and only if <math>p(x)</math> has y-intercept <math>r</math>, where <math>r</math> is some constant. What is the value of <math>r + k</math>? (Source: [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi])
+*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])

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Difference between revisions of "Titu's Lemma"

Latest revision as of 19:10, 29 April 2024

Contents

Examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Problems

Introductory

Intermediate

Olympiad