**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 \(x, y, z\)

\(\frac{2}{x + y}+\frac{2}{y + z}+\frac{2}{z + x} \ge \frac{9}{x + y + z}\)

## 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 – Dinesh Chavan · 3 years, 7 months ago

Log in to reply

– Priyanshu Mishra · 1 year, 10 months ago

Is it not possible by setting \(x+y+z=1\) and then applying the lemma?Log in to reply

– Eddie The Head · 3 years, 7 months ago

I multiplied up and down by 2...follows the same steps though.....Cheers!Log in to reply

– Zi Song Yeoh · 3 years, 7 months ago

There is a minor typo at the end of your solution. The \(3\) should be inside the bracket.Log in to reply

– Dinesh Chavan · 3 years, 7 months ago

Got it.Sorry!Log in to reply

– Megh Choksi · 2 years, 6 months ago

Very direct application questionLog in to reply

Stay tuned for Part 2! – Zi Song Yeoh · 3 years, 7 months ago

Log in to reply

– Eddie The Head · 3 years, 7 months ago

Really nice post...Part II please!!!Log in to reply

other method for 1)

\( A.M \geq H.M\)

\( s = a + b + c\)

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

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

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

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

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

\( \dfrac{a}{b + c} + \dfrac{b}{a + c} + \dfrac{c}{b + c} \geq \dfrac{3}{2}\) – Megh Choksi · 2 years, 6 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. – Happy Melodies · 3 years, 3 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) – Radovan Švarc · 3 years, 7 months ago

Log in to reply

– Michael Tang · 3 years, 7 months ago

Many also call it Cauchy "in fractional form," which is similar to your last name.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 :) – Nihar Mahajan · 1 year, 9 months ago

Log in to reply

very nice way to prove it – Niaz Ghumro · 2 years, 7 months ago

Log in to reply

Can you add this to Titu's Lemma? Thanks @Zi Song Yeoh – Calvin Lin Staff · 2 years, 9 months ago

Log in to reply

Actually, this is a special case of Holder's Inequality – Zi Song Yeoh · 3 years, 7 months ago

Log in to reply