Titu's Lemma

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 $p(x) = x^3 + bx^2 + cx + d$ for some real $b$, $c$, and $d$. There exists some minimum integer $k>1$ such that the equation $h = \sqrt{\frac{x^2yz + 3x^2y + 2x^2z + 6x^2 + xy^2z + 3xy^2 + y^2z + 3y^2 + xyz^2 + 2xz^2 + yz^2 + 2z^2}{xyz + 3xy + 2xz + 6x + yz + 3y + 2z + 6}}$ has some solution in $x$, $y$, and $z$ when $h=k$ and no solutions when $h<k$. This is the case if and only if $p(x)$ has y-intercept $r$, where $r$ is some constant. What is the value of $r + k$? (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)