Titu's Lemma

Titu's Lemma is a consequence of Cauchy-Schwarz inequality, but this special form helps tackle a lot of optimization problems involving squares in the numerator, immediately. For two sequences \(a_1,...,a_n\) and \(b_1,...,b_n\) of positive reals, it is stated as

\[\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}\ge \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n} .\]

This lemma is also known as Engel's form of Cauchy-Schwarz inequality.

Since Titu's Lemma is a consequence of Cauchy-Schwarz inequality, first we recall Cauchy-Schwarz inequality which is stated as follows:

For two sequences \(a_1,...,a_n\) and \(b_1,...,b_n\) of positive reals the Cauchy-Schwarz inequality states that

\[\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge \left(a_1b_1+a_2b_2+\cdots+a_nb_n\right)^2 \ , \]

where the equality is met if and only if \(\dfrac{a_i/\sqrt{b_i}}{\sqrt{b_i}}=\dfrac{a_i}{b_i}\) is constant for all \(i\).

We replace \(a_i\to \dfrac{a_i}{\sqrt{b_i}}\) and \(b_i\to \sqrt{b_i}\) for \(1\le i\le n\), then the above inequality reads

\[\left(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots +\dfrac{a_n^2}{b_n}\right)\left(b_1+b_2+\cdots+b_n\right)\ge \left(a_1+a_2+\cdots+a_n\right)^2,\]

which is equivalent to

\[\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}\ge \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}. \qquad \text{(Titu's Lemma)}\]

So any expression containing squares in the numerator, and any term in the denominator can be cleared away leading to one single fraction, applying this awesome lemma. The following examples will help in learning the applications of Titu's Lemma.

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 by Titu's lemma

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

which is exactly what we wanted. \(_\square\)

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. So applying Titu's lemma, we get

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

Again by Titu's lemma, 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

\[\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{\left(\sqrt{2}\right)^{2}}{x+y}+\frac{\left(\sqrt{2}\right)^{2}}{y+z}+\frac{\left(\sqrt{2}\right)^{2}}{z+x}.\]

Then by Titu's Lemma,

\[\frac{\left(\sqrt{2}\right)^{2}}{x+y}+\frac{\left(\sqrt{2}\right)^{2}}{y+z}+\frac{\left(\sqrt{2}\right)^{2}}{z+x}≥\frac{(3\sqrt{2})^{2}}{2(x+y+z)}=\frac{9}{x+y+z}. \ _\square\]

Nesbitt's Inequality:

For \(a,b,c\in\mathbb{R^{+}}\), prove that \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}.\]

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Observe that

\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{ac+bc}\]

and that

\[\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{ac+bc}≥\frac{(a+b+c)^{2}}{2(ab+bc+ca)}.\]

Then it suffices to prove

\[\frac{(a+b+c)^{2}}{2(ab+bc+ca)}≥\frac{3}{2}.\]

A short computation shows that this is equivalent to \(a^{2}+b^{2}+c^{2}≥ab+bc+ca\), which is true for any positive reals \(a,b,c\). \(_\square\)

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

\[8(a^{4}+b^{4})≥(a+b)^{4}.\]

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We apply Titu's Lemma twice to show \[a^{4}+b^{4}=\dfrac{a^{4}}{1}+\dfrac{b^{4}}{1}≥\dfrac{\left(a^{2}+b^{2}\right)^{2}}{2}≥\dfrac{\left(\frac{(a+b)^{2}}{2}\right)^{2}}{2}=\dfrac{(a+b)^{4}}{8}. \ _\square\]

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 given expression 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 by Titu's Lemma is \(≥\dfrac{(2a+2b+2c)^{2}}{4(a+b+c)}\) and we are done. \(_\square\)

Here is the generalized form of Titu's Lemma:

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

Then by simplifying we get

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

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

When we put \(m=2\) in the above inequality, we get

\[\frac{x_1^2}{a_1}+\frac{x_2^2}{a_2}+\cdots+\frac{x_n^2}{a_1}\ge\frac{(x_1+x_2+\cdots+x_n)^2}{(a_1+a_2+\cdots+a_n)} \ , \]

which is Titu's Lemma itself.

This section is motivated to boost up the problem solving skills on the notion of Titu's lemma. We start with some examples followed by several problems to try.

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 given expression as \(\dfrac{\frac{1}{a^{2}}}{ab+ac}+\dfrac{\frac{1}{b^{2}}}{ba+bc}+\dfrac{\frac{1}{c^{2}}}{ca+cb}\).

By Titu's Lemma, this is \(≥\dfrac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{2(ab+bc+ca)}\).

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

Hence, the given expression 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:

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

Now, applying Titu's Lemma gives

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

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 \(\dfrac{x^3}{(x+y)^2}+\dfrac{y^3}{(y+z)^2}+\dfrac{z^3}{(z+x)^2}\ge \dfrac14\). Thus by applying the above inequality in the LHS, we get

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

Hence proved. \(_\square\).

Try the following examples.

• Cauchy-Schwarz Inequality

• Holder's Inequality

• Rearrangement Inequality

• Jensen's Inequality