# Titu's Lemma (Part 2)

In part 1, we saw what Titu's Lemma is and one example of using it. Now, let's see more examples of using it.

$2.$ (TOT 1998) Prove that for any positive real numbers $a, b, c$,

$\frac{a^3}{a^2 + ab + b^2} + \frac{b^3}{b^2 + bc + c^2} + \frac{c^3}{a^2 + ac + c^2} \ge \frac{a + b + c}{3}$

Solution :

Again, we want to make the numerators on the left hand side be squares.

So, we again multiply the numerators and denominators of the fractions by $a, b, c$ respectively.

By Titu's Lemma,

$\frac{a^4}{a(a^2 + ab + b^2)} + \frac{b^4}{b(b^2 + bc + c^2)} + \frac{c^4}{c(a^2 + ac + c^2)}$

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

But the denominator is equivalent to $(a + b + c)(a^2 + b^2 + c^2)$.

Thus, we the inequality becomes

$\frac{a^4}{a(a^2 + ab + b^2)} + \frac{b^4}{b(b^2 + bc + c^2)} + \frac{c^4}{c(a^2 + ac + c^2)} \ge \frac{a^2 + b^2 + c^2}{a + b + c}$

Finally, we have to prove that $\frac{a^2 + b^2 + c^2}{a + b + c} \ge \frac{a + b + c}{3}$, which is true. (The proof is similar to the last problem.)

$3.$ (IMO 1995) Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that

$\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(b + a)} \ge \frac{3}{2}$.

Solution : Direct application of Titu's Lemma is unfortunately doomed to failure. We divide the numerator and denominator by $a^2, b^2$ and $c^2$ respectively.

Now, we have by Titu's Lemma,

$\frac{1}{a^3(b + c)} + \frac{1}{b^3(a + c)} + \frac{1}{c^3(b + a)} = \frac{\frac{1}{a^2}}{a(b + c)} + \frac{\frac{1}{b^2}}{b(a + c)} + \frac{\frac{1}{c^2}}{c(b + a)}$

$\ge \frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{2(ab + bc + ca)} = \frac{(ab + bc + ca)^2}{2(ab + bc + ca)}$

$\ge \frac{ab + bc + ca}{2} \ge \frac{3\sqrt[3]{(abc)^2}}{2} = \frac{3}{2}$

where the last inequality holds by AM-GM.

This proves the inequality.

Problems

$2.$ Prove that for all positive real numbers $a, b, c, d$

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

$3.$ Problem link

$4.$ (IMO 2005)

Prove that for all positive real numbers $x, y, z$ such that $xyz \ge 1$, then

$\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} + \frac{x^2 + y^2 + z^2}{y^5 + x^2 + z^2} + \frac{x^2 + y^2 + z^2}{z^5 + y^2 + x^2} \le 3$

Note : Problem 2 uses the same idea as Example 2, but Problem 4 is a bit harder, for a few reasons :

1. There are no square terms on the numerators.

2. The inequality sign is less than or equal to.

Note by Zi Song Yeoh
6 years, 5 months ago

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Will there be a part 3 with ore problems??

- 6 years, 5 months ago

Great posts! Looking forwards to more inequality posts, as inequalities are my worst subject.

- 6 years, 5 months ago

Hello @Daniel Chiu ! Check my posts for inequalities :D

- 5 years, 10 months ago

Small typo: on the first problem, after "by titu's lemma", the b and c should be to the 4th power.

- 6 years, 5 months ago

Also, Thank you for such great posts! I had no idea at all how to solve complex inequalities before, but with this tool under my belt I can do a lot more than I previously ever thought I could do.

Previously, I had known Cauchy-Schwarz, but I had no idea how to effectively apply it. Can you please make a topic on Cauchy Schwarz? Thanks!

And a solution outline for 1:

Multiply numerator and denominator of each fraction in LHS by a,b,c,d respectively. Apply Titu's Lemma, then clear denominators and section out the variables to form multiple sums of squares. This is always mor than 0, so we have proved it.

- 6 years, 5 months ago

- 6 years, 5 months ago

Would you be kind enough to give me the link? Thanks.

- 6 years, 5 months ago

- 6 years, 5 months ago

Edited. Thanks for pointing out.

- 6 years, 5 months ago

Hints on Prob 4??? :(

- 6 years, 5 months ago

There is a 2 step solution to Problem 4, which won the Special Prize at IMO'05.

Staff - 6 years, 5 months ago

Whoa!!A little hint?? :p

- 6 years, 5 months ago

It involves finding an identity which is almost impossible to find.

- 6 years, 5 months ago

You may try with another method other then the one I have posted. I got one more.

- 6 years, 5 months ago

Hi, Yesterday I saw the fourth problem and it really trolled me I tried for about 5 hours and ended up with this solution.Hope it works. We have to prove the following

$\frac{x^2+y^2+z^2}{x^5+y^2+z^2}+\frac{x^2+y^2+z^2}{y^5+x^2+z^2}+\frac{x^2+y^2+z^2}{z^5+y^2+x^2}≤3$ Which is equivalent to

$(x^2+y^2+z^2)(\frac{1}{x^5+y^2+z^2}+\frac{1}{y^5+x^2+z^2}+\frac{1}{z^5+y^2+x^2})≤3$ Which is same as

$\frac{1}{x^5+y^2+z^2}+\frac{1}{y^5+x^2+z^2}+\frac{1}{z^5+y^2+x^2}≤\frac{3}{x^2+y^2+z^2}$

The key step to prove the inequality is the following step. We have by Cauchy-Schwarz inequality :(https://brilliant.org/discussions/thread/elementary-techniques-used-in-the-imo-internatio-V/)

By using the inequality, we get $(x^5+y^2+z^2)(yz+y^2+z^2)≥(\sqrt{x^5yz}+y^2+z^2)^2$ And by using the fact that $xyz≥1$ we get $(x^5+y^2+z^2)(yz+y^2+z^2)≥(\sqrt{x^5yz}+y^2+z^2)^2≥(x^2+y^2+z^2)^2$ Now by using the above fact the problem becomes very simple as $\frac{1}{x^5+y^2+z^2}≤\frac{yz+y^2+z^2}{(x^2+y^2+z^2)^2}≤\frac{\frac{z^2+y^2}{2}+y^2+z^2}{(x^2+y^2+z^2)^2}$ ........ By AM-GM inequality..

Using the same method for $y$ and $z$, we get; $\frac{1}{y^5+x^2+z^2}≤\frac{\frac{x^2+z^2}{2}+x^2+z^2}{(x^2+y^2+z^2)^2}$ and

$\frac{1}{z^5+y^2+x^2}≤\frac{\frac{x^2+y^2}{2}+x^2+y^2}{(x^2+y^2+z^2)^2}$

Now adding the above three results gives the expected result. And Now I really want to thank Zi Song for posting such wonderful posts on inequalities.I love inequalities the most and I am looking forward for part 3 of your post Very good and please keep posting such good notes .Thanks.

- 6 years, 5 months ago

Really nice....I also spent a bit of time on this problem....

- 6 years, 5 months ago

Thanks

- 6 years, 5 months ago

how can we solve question no 2

- 4 years, 4 months ago

For problem 4, $(x^5+y^2+z^2)\left(\frac1x + y^2 + z^2\right) \geq (x^2 + y^2 + z^2)^2$ $\iff \frac{x^2+y^2+z^2}{x^5+y^2+z^2} \leq \frac{\frac1x + y^2 + z^2}{x^2+y^2+z^2} \leq \frac{yz + y^2 + z^2}{x^2+y^2+z^2}$ Adding similar inequalities, $\sum_{cyc} \frac{x^2+y^2+z^2}{x^5+y^2+z^2}\leq \frac{\sum_{cyc} (yz + y^2 +z ^2)}{x^2+y^2+z^2}$ $= \frac{2(x^2+y^2+z^2) + (xy+yz+zx)}{x^2+y^2+z^2} = 2 + \frac{xy+yz+zx}{x^2+y^2+z^2} \leq 3$ The final equality follows, since, $x^2 + y^2 + z^2 \geq xy + yz +zx \iff \frac12[(x-y)^2 + (y-z)^2 + (z-x)^2] \geq 0.$

- 2 years ago

For problem 2, $\sum_{cyc} \frac{a}{b+2c+3d} = \sum_{cyc} \frac{a^2}{ab+2ca+3da} \geq \frac{(a+b+c+d)^2}{\sum_{cyc} (ab+2ac+3ad)}$ $= \frac{(a+b+c+d)^2}{4(ab+bc+cd+da+ca+bd)} = \frac{a^2+b^2+c^2+d^2}{4(ab+bc+cd+da+ca+bd)} + \frac12$ So, it suffices to prove that $\frac{a^2+b^2+c^2+d^2}{4(ab+bc+cd+da+ca+bd)} \geq \frac16 \iff 3(a^2+b^2+c^2+d^2) \geq 2(ab+bc+cd+da+ca+bd)$ $\iff (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + (b-d)^2 + (a-c)^2 \geq 0$

- 2 years ago