# 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
7 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

Sort by:

Top Newest

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, $\frac{(\sqrt{2})^2}{x+y}+\frac{(\sqrt{2})^2}{z+y}+\frac{(\sqrt{2})^2}{x+z} \geq \frac{(3\sqrt{2})^2}{2(x+y+z)}=\frac{9}{x+y+z}$

I liked your thread for inequality. Looking forward for next problems

- 7 years, 4 months ago

Log in to reply

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

- 7 years, 4 months ago

Log in to reply

Got it.Sorry!

- 7 years, 3 months ago

Log in to reply

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

- 7 years, 4 months ago

Log in to reply

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

- 5 years, 6 months ago

Log in to reply

Very direct application question

- 6 years, 2 months ago

Log in to reply

Stay tuned for Part 2!

- 7 years, 4 months ago

Log in to reply

Really nice post...Part II please!!!

- 7 years, 4 months ago

Log in to reply

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.

- 7 years ago

Log in to reply

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, 5 months ago

Log in to reply

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)

- 7 years, 3 months ago

Log in to reply

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

- 7 years, 3 months ago

Log in to reply

Some even call it Cauchy in Engel Form

- 3 years, 4 months ago

Log in to reply

other method for 1)

$A.M \geq H.M$

$s = a + b + c$

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

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

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

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

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

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

- 6 years, 2 months ago

Log in to reply

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

- 7 years, 4 months ago

Log in to reply

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

Staff - 6 years, 6 months ago

Log in to reply

very nice way to prove it

- 6 years, 4 months ago

Log in to reply

@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, 6 months ago

Log in to reply

Wishful thinking does motivate us indeed.

Also nice notes

- 3 years, 7 months ago

Log in to reply

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

- 2 years, 11 months ago

Log in to reply

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, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...