# Titu's Lemma (Part 1)

Titu's lemma

The Inequality

Let $x_1, x_2, ..., x_n$ be real numbers and $y_1, y_2, ..., y_n$ be positive real numbers.

Then, the following inequality holds,

$\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + ... + \frac{x_n^2}{y_n} \ge \frac{(x_1 + x_2 + ... + x_n)^2}{y_1 + y_2 + ... + y_n}$

Why is this true? Actually, this inequality follows from the Cauchy-Schwarz Inequality.

$(\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + ... + \frac{x_n^2}{y_n})(y_1 + y_2 + ... + y_n) \ge (x_1 + x_2 + ... + x_n)^2$

Examples

$1.$ Prove Nesbitt Inequality :

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

Solution :

We can't see any squares terms on the numerators, so wishful thinking motivates us to create them.

How can we create the square terms? Squaring the whole left hand side is very messy, and a much simpler way is to multiply the numerators and denominators of the fractions by $a, b, c$ respectively.

Now, the way to proceed is clear, as by Titu's Lemma we get

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

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

Now, we just have to prove that $2(a + b + c)^2 \ge 6(ab + bc + ca)$, which can be rewritten as

$(a + b + c)^2 \ge 3(ab + bc + ca)$, which can be rewritten as $a^2 + b^2 + c^2 \ge ab + bc + ca$,

The last inequality follows from $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (a - c)^2] \ge 0$.

Problems

$1.$ Prove that for all positive real numbers $x, y, z$

$\frac{2}{x + y}+\frac{2}{y + z}+\frac{2}{z + x} \ge \frac{9}{x + y + z}$

Note by Zi Song Yeoh
6 years, 11 months ago

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Hi, Here is your answer We can write $\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}$ as $\frac{(\sqrt{2})^2}{x+y}+\frac{(\sqrt{2})^2}{z+y}+\frac{(\sqrt{2})^2}{x+z}$ So as to use The above mentioned inequality. So after applying the inequality, we get, $\frac{(\sqrt{2})^2}{x+y}+\frac{(\sqrt{2})^2}{z+y}+\frac{(\sqrt{2})^2}{x+z} \geq \frac{(3\sqrt{2})^2}{2(x+y+z)}=\frac{9}{x+y+z}$

- 6 years, 11 months ago

There is a minor typo at the end of your solution. The $3$ should be inside the bracket.

- 6 years, 11 months ago

Got it.Sorry!

- 6 years, 11 months ago

I multiplied up and down by 2...follows the same steps though.....Cheers!

- 6 years, 11 months ago

Is it not possible by setting $x+y+z=1$ and then applying the lemma?

- 5 years, 2 months ago

Very direct application question

- 5 years, 10 months ago

Stay tuned for Part 2!

- 6 years, 11 months ago

- 6 years, 11 months ago

I did a similar method but slightly more intuitive..

First observe $9 = 3^2$, therefore, we can try to create the sum of the numerators of the fractions in LHS to be $3$.

Also, observe that there is a common multiple of $2$ in the numerator in LHS, so divide the LHS and RHS with $2$ to get: $\frac{1}{x+y} + \frac{1}{y+z}+\frac{1}{z+x}\geq \frac{9}{2(x+y+z)}$

Then using Titu's Lemma, we get $\frac{1}{x+y} + \frac{1}{y+z}+\frac{1}{z+x}\geq \frac{(1+1+1)^2}{2(x+y+z)} = \frac{9}{2(x+y+z)}$. We are done.

- 6 years, 7 months ago

This is NOT Titu's lemma. This inequality was published in 1997 in russian language in journal KVANT by Nairi Sedrakyan, Sedrakyan not only wrote the inequality in this form, but he has shown how this inequality can be used to prove different inequalities and called his article About applications of one useful inequality. In russian speaking countries it is called Sedrakyan's inequality. Here is the link of the first page of that article http://kvant.mccme.ru/pdf/1997/02/42.pdf But in english speaking countries this inequality sometimes is called Engel's form or Titu's Lemma. Because Arthur Engel has included it in his book published in 1998 and Titu Andreescu has included it in his book published in 2003. Both authors were familiar with russian language mathematics literature and Sedrakyan's works, but none of them cited the original source of the work. Moreover, Sedrakyan devotes a whole separate chapter to the applications of this inequality in his book Inequalities. Methods of proving published in 2002 in russian.

- 3 years, 1 month ago

Pretty interesting is name of this. Except Titu's lemma (sometimes written as T2 lemma), you can also find it as SQ lemma (like ...escu), or (at least in Czech and Slovak republic) as CS-fractionfighter (CS stands for Cauchy-Schwartz)

- 6 years, 11 months ago

Many also call it Cauchy "in fractional form," which is similar to your last name.

- 6 years, 11 months ago

Some even call it Cauchy in Engel Form

- 3 years ago

other method for 1)

$A.M \geq H.M$

$s = a + b + c$

$\dfrac{a}{s - a} + \dfrac{b}{s - b} + \dfrac{c}{s - c} = \dfrac{s}{s - a} - 1 + \dfrac{s}{s - b} - 1 + \dfrac{s}{s - c} - 1$

$\dfrac{\dfrac{s}{s - a} + \dfrac{s}{s - b} + \dfrac{s}{s - c}}{3} \geq \dfrac{3}{\dfrac{s - a + s - b + s - c}{s}}$

$\dfrac{\dfrac{s}{s - a} + \dfrac{s}{s - b} + \dfrac{s}{s - c}}{3} \geq \dfrac{3}{2}$

$\dfrac{\dfrac{s}{s - a} - 1 + \dfrac{s}{s - b} - 1 + \dfrac{s}{s - c} - 1 }{3} \geq \dfrac{1}{2}$

$\dfrac{s}{s - a} - 1 + \dfrac{s}{s - b} - 1 + \dfrac{s}{s - c} - 1 \geq \dfrac{3}{2}$

$\dfrac{a}{b + c} + \dfrac{b}{a + c} + \dfrac{c}{b + c} \geq \dfrac{3}{2}$

- 5 years, 10 months ago

Actually, this is a special case of Holder's Inequality

- 6 years, 11 months ago

Can you add this to Titu's Lemma? Thanks @Zi Song Yeoh

Staff - 6 years, 1 month ago

very nice way to prove it

- 5 years, 11 months ago

@Zi Song Yeoh @Calvin Lin Thanks , this note was quite useful. For saying thank you I created this problem : click here. Hope you enjoy it :)

- 5 years, 2 months ago

Wishful thinking does motivate us indeed.

Also nice notes

- 3 years, 3 months ago

$2\left(\sum_{cyc} \frac1{x+y}\right) \geq 2\cdot \frac{(1+1+1)^2}{2(x+y+z)} = \frac9{x+y+z}$

- 2 years, 7 months ago

Cauchy Schwarz inequality states, [(x+y)+(y+z)+(z+x)].[1/(x+y)+1/(y+z)+1/(z+x)]≥(1+1+1)^2...........(£)

Hence the rearrangement of the above proves the requirement

Also Eqn. (£) can be produced by AM-GM

- 2 years, 6 months ago