**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 $x, y, z$

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHi, 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

Log in to reply

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

Log in to reply

Got it.Sorry!

Log in to reply

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

Log in to reply

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

Log in to reply

Very direct application question

Log in to reply

Stay tuned for Part 2!

Log in to reply

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

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)

Log in to reply

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

Log in to reply

Some even call it Cauchy in Engel Form

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.

Log in to reply

This is

NOTTitu's lemma. This inequality was published in 1997 in russian language in journal KVANT byNairi 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 articleAbout applications of one useful inequality. In russian speaking countries it is calledSedrakyan'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 calledEngel's formorTitu's Lemma. BecauseArthur Engelhas included it in his book published in 1998 andTitu Andreescuhas 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 devotesa whole separate chapterto the applications of this inequality in his bookInequalities. Methods of provingpublished in 2002 in russian.Log in to reply

other method for 1)

$A.M \geq H.M$

$s = a + b + c$

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

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

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

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

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

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

Log in to reply

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

Log in to reply

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

Log in to reply

very nice way to prove it

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 :)

Log in to reply

Wishful thinking does motivate us indeed.

Also nice notes

Log in to reply

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

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

Log in to reply