Titu's Lemma

Titu's lemma (also known as T2 Lemma, Engel's form, or Sedrakyan's inequality) states that for positive reals $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$, we have

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

It is a direct consequence of Cauchy-Schwarz inequality. This form is especially helpful when the inequality involves fractions where the numerator is a perfect square.

It is obtained by applying the substitution $a_i= \frac{x_i}{ \sqrt{y_i} }$ and $b_i = \sqrt{y_i}$ into the Cauchy-Schwarz inequality. Equality holds if and only if $a_i = k b_i$ for a non-zero real constant $k$.

It then becomes

$\begin{aligned} \left( \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } \right) (y_1 + y_2 + \cdots + y_n ) &\geq ( x_1 + x_2 + \cdots + x_n )^2 \\ \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } &\geq \frac{ ( x_1 + x_2 + \cdots + x_n )^2 }{ (y_1 + y_2 + \cdots + y_n ) }. \end{aligned}$

For positive reals $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$, we have

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

It is obtained by applying the substitution $a_i= \frac{x_i}{ \sqrt{y_i} }$ and $b_i = \sqrt{y_i}$ into the Cauchy-Schwarz inequality.

It then becomes

$\begin{aligned} \left( \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } \right) (y_1 + y_2 + \cdots + y_n ) &\geq ( x_1 + x_2 + \cdots + x_n )^2 \\ \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } &\geq \frac{ ( x_1 + x_2 + \cdots + x_n )^2 }{ (y_1 + y_2 + \cdots + y_n ) }. \ _\square \end{aligned}$

Prove that for positive reals $a,b,c$ summing up to $1$, we have

$\dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 98.$

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The hypothesis implies $a+b+c=1$. We have square terms in the numerator, so

$\begin{aligned} \dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c} &=& \dfrac{1}{a+b}+\dfrac{4^2}{c}+\dfrac{9^2}{a+b+c} \\ \\ &\ge &\dfrac{(1+4+9)^2}{(a+b)+c+(a+b+c)} \\ \\ & = & \dfrac{(14)^2}{2(a+b+c)} = \dfrac{196}{2} = 98, \end{aligned}$

which is exactly what we wanted. $_\square$

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Note that the equality condition $a+b = k, c = 16k, a+b+c = 81k$ cannot be satisfied.

In fact, we have the stronger statement:

$\dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 106.$

Can you prove this using Titu's lemma?

Over all triplets of positive reals $x,y,z$ satisfying $x+y+z=3$, find the minimum value of

$f(x,y,z)=\dfrac{yz+4zx+9xy}{xyz}.$

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Dividing out simplifies $f(x,y,z)$ into $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$. Now we have square terms on the numerator. We get

$f(x,y,z)=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z} \ge \dfrac{(1+2+3)^2}{x+y+z} = \dfrac{36}{3}=12.$

Equality holds when

$\dfrac 1 x = \dfrac 2 y = \dfrac 3 z~\Longleftrightarrow ~ y=2x, ~z=\dfrac{3y}{2}=3x,$

and solving for $x$ using the constraint $x+y+z=3$, we have equality at $x=\frac{1}{2}$, $y=1$, $z=\frac{3}{2}$. We check that equality indeed occurs.

Therefore, the minimum value is 12. $_\square$

Let $x,y,z\in\mathbb{R^{+}}$. Prove that $\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}≥\frac{9}{x+y+z}.$

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Observe that

$\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}.$

Then,

$\frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}≥\frac{\big(3\sqrt{2}\big)^{2}}{2(x+y+z)}=\frac{9}{x+y+z}. \ _\square$

Let $a,b\in\mathbb{R^{+}}$. Prove that

$8\big(a^{4}+b^{4}\big)\geq(a+b)^{4}.$

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We have

$a^{4}+b^{4}=\dfrac{a^{4}}{1}+\dfrac{b^{4}}{1}\geq\dfrac{\left(a^{2}+b^{2}\right)^{2}}{2}\geq\dfrac{\Big(\frac{(a+b)^{2}}{2}\Big)^{2}}{2}=\dfrac{(a+b)^{4}}{8}. \ _\square$

For $a,b,c\in\mathbb{R^{+}}$, prove that

$\frac{a^{2}+b^{2}}{a+b}+\frac{b^{2}+c^{2}}{b+c}+\frac{c^{2}+a^{2}}{c+a}≥a+b+c.$

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Write the left side of the given inequality as

$\dfrac{a^{2}}{a+b}+\dfrac{b^{2}}{a+b}+\dfrac{b^{2}}{b+c}+\dfrac{a^{2}}{c+a}+\dfrac{c^{2}}{a+c}+\dfrac{c^{2}}{c+b},$

which is $≥\frac{(2a+2b+2c)^{2}}{4(a+b+c)},$ so we are done. $_\square$

Now you can try to prove the Nesbitt's inequality.

Nesbitt's Inequality

Let $a, b, c \in \mathbb{R}^+$. Show that

$\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \ge \frac{3}{2}.$

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SHOW SOLUTION

The numerator currently isn't a perfect square. Let's make it a perfect square:

$\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} = \frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } .$

We can then apply Titu's lemma to obtain

$\frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } \geq \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } .$

Now,

$\begin{array} { c r c l } & \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } & \geq & \frac{3}{2} \\ \Leftrightarrow & (a+b+c)^2 & \geq & 3 (ab+bc+ca) \\ \Leftrightarrow & a^2 + b^2 + c^2 & \geq & ab + bc + ca \\ \Leftrightarrow & (a-b)^2+(b-c)^2+(c-a)^2 & \geq & 0. \end{array}$

Note that if we tried to apply Titu's lemma directly, we end up with

$\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \geq \frac{ \big( \sqrt{a} + \sqrt{b} + \sqrt{c} \big)^2 } { 2 ( a + b + c ) }.$

Then, the right-hand side actually has a maximum value of $\frac{3}{2}$, so we cannot proceed any further. $_\square$

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Another approach is to apply Titu's lemma to show that:

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

Hence,

$\frac{a+b+c}{ b+c} + \frac{a+b+c}{ c+a} + \frac{a+b+c}{a+b} \geq \frac{9}{2}.$

Hence,

$\frac{a}{ b+c} + \frac{b}{ c+a} + \frac{c}{a+b} \geq \frac{9}{2} - 3 = \frac{3}{2} .$

Here is the generalized form:

For any positive integer $m$, we have $\frac{x_1^m}{a_1^{m-1}}+\frac{x_2^m}{a_2^{m-1}}+\cdots+\frac{x_n^m}{a_n^{m-1}}\ge\frac{(x_1+x_2+\cdots+x_n)^m}{(a_1+a_2+\cdots+a_n)^{m-1}}.$

To learn how we reached this generalized form, follow the process below.

Let $\displaystyle \bigg(\sum_{i=1}^n\sqrt[\frac{m}{m-1}]{a_i}^{\frac{m}{m-1}}\bigg)^{\frac{1}{\ \frac{m}{m-1}\ }}\left(\sum_{i=1}^n\dfrac{x_i^m}{\left(\frac{a_i}{\sqrt[m]{a_i}}\right)^m}\right)^{\frac{1}{m}} \ge \sum_{i=1}^n x_i,$ which is true by Hölder's inequality.

Then by simplifying, we get

$\begin{aligned}\displaystyle \bigg(\sum_{i=1}^n a_i\bigg)^{\frac{m-1}{m}}\bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)^{\frac{1}{m}}&\ge \sum_{i=1}^n x_i \\\bigg(\sum_{i=1}^n a_i\bigg)^{m-1}\bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)&\ge \bigg(\sum_{i=1}^n x_i\bigg)^m \\ \bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)&\ge \dfrac{\bigg(\sum_{i=1}^n x_i\bigg)^m}{\bigg(\sum_{i=1}^n a_i\bigg)^{m-1}}.\end{aligned}$

Thus we conclude that ${\frac{x_1^m}{a_1^{m-1}}+\frac{x_2^m}{a_2^{m-1}}+\cdots+\frac{x_n^m}{a_n^{m-1}}\ge\frac{(x_1+x_2+\cdots+x_n)^m}{(a_1+a_2+\cdots+a_n)^{m-1}}}$ for $m \ge 2.$

When we put $m=2$ in the above inequality, we get

$\frac{x_1^2}{a_1}+\frac{x_2^2}{a_2}+\cdots+\frac{x_n^2}{a_n}\ge\frac{(x_1+x_2+\cdots+x_n)^2}{(a_1+a_2+\cdots+a_n)}.$

We start with some examples followed by several problems to try.

[IMO/1995] Given $a,b,c\in\mathbb{R^{+}} \text{ and } abc=1$, prove that

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

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We write the left side of the given inequality as $\dfrac{\frac{1}{a^{2}}}{ab+ac}+\dfrac{\frac{1}{b^{2}}}{ba+bc}+\dfrac{\frac{1}{c^{2}}}{ca+cb}$.

This is $\ge \dfrac{\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)^{2}}{2(ab+bc+ca)}$.

Since $abc=1$, it follows that $\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)^{2}=(ab+bc+ca)^{2}$.

Hence, the left-hand side of the given inequality is $≥\dfrac{(ab+bc+ca)}{2}≥3\dfrac{\sqrt[3]{(abc)^{2}}}{2}=\dfrac{3}{2}$. $_\square$

For $a,b,c\in\mathbb{R^{+}}$, prove that

$\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}≤1.$

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Let's modify the inequality that we have to prove as follows:

$\begin{aligned} \left(\frac{a}{2a+b}-\frac{1}{2}\right)+\left(\frac{b}{2b+c}-\frac{1}{2}\right)+\left(\frac{c}{2c+a}-\frac{1}{2}\right) &\leq 1-\frac{3}{2} \\ \frac{2a-(2a+b)}{2(2a+b)}+\frac{2b-(2b+c)}{2(2b+c)}+\frac{2c-(2c+a)}{2(2c+a)} &\leq -\frac{1}{2} \\ -\frac{1}{2}\left( \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\right) &\leq -\frac{1}{2} \\ \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a} &\geq 1. \end{aligned}$

Then we have

$\frac{b^2}{2ab+b^2}+\frac{c^2}{2bc+c^2}+\frac{a^2}{2ca+a^2} \geq \frac{(a+b+c)^2}{a^2+b^2+c^2+2(ab+bc+ca)}=1. \ _\square$

For the positive real numbers $x, y,$ and $z,$ where $x+y+z=1,$ prove that

$\displaystyle \sum_{cyc} \frac{x^3}{(x+y)^2} \ge \dfrac14.$

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What we should actually prove is that $\frac{x^3}{(x+y)^2}+\frac{y^3}{(y+z)^2}+\frac{z^3}{(z+x)^2}\ge \frac14$. Thus by applying the above inequality on the LHS, we get

$\begin{aligned}\dfrac{x^3}{(x+y)^2}+\dfrac{y^3}{(y+z)^2}+\dfrac{z^3}{(z+x)^2}&\ge \dfrac{(x+y+z)^3}{(x+y+y+z+z+x)^2} \\&=\dfrac{(x+y+z)^3}{4(x+y+z)^2} \\&=\dfrac{1}{4}.\end{aligned}$

Hence proved. $_\square$

Try the following examples.

• Cauchy-Schwarz Inequality

• Holder's Inequality

• Rearrangement Inequality

• Jensen's Inequality

• Nesbitt's Inequality