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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
3 years, 5 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, \(\frac{(\sqrt{2})^2}{x+y}+\frac{(\sqrt{2})^2}{z+y}+\frac{(\sqrt{2})^2}{x+z} \geq \frac{(3\sqrt{2})^2}{2(x+y+z)}=\frac{9}{x+y+z}\)

I liked your thread for inequality. Looking forward for next problems Dinesh Chavan · 3 years, 5 months ago

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@Dinesh Chavan Is it not possible by setting \(x+y+z=1\) and then applying the lemma? Priyanshu Mishra · 1 year, 8 months ago

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@Dinesh Chavan I multiplied up and down by 2...follows the same steps though.....Cheers! Eddie The Head · 3 years, 5 months ago

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@Dinesh Chavan There is a minor typo at the end of your solution. The \(3\) should be inside the bracket. Zi Song Yeoh · 3 years, 5 months ago

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@Zi Song Yeoh Got it.Sorry! Dinesh Chavan · 3 years, 5 months ago

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@Dinesh Chavan Very direct application question Megh Choksi · 2 years, 4 months ago

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Stay tuned for Part 2! Zi Song Yeoh · 3 years, 5 months ago

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@Zi Song Yeoh Really nice post...Part II please!!! Eddie The Head · 3 years, 5 months ago

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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, 4 months ago

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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, 1 month ago

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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, 5 months ago

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@Radovan Švarc Many also call it Cauchy "in fractional form," which is similar to your last name. Michael Tang · 3 years, 5 months ago

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@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, 8 months ago

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very nice way to prove it Niaz Ghumro · 2 years, 5 months ago

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Can you add this to Titu's Lemma? Thanks @Zi Song Yeoh Calvin Lin Staff · 2 years, 7 months ago

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Actually, this is a special case of Holder's Inequality Zi Song Yeoh · 3 years, 5 months ago

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