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.

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

a3a2+ab+b2+b3b2+bc+c2+c3a2+ac+c2a+b+c3\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,ca, b, c respectively.

By Titu's Lemma, 

a4a(a2+ab+b2)+b4b(b2+bc+c2)+c4c(a2+ac+c2)\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)}

(a2+b2+c2)2a(a2+ab+b2)+b(b2+bc+c2)+c(a2+ac+c2)\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)(a2+b2+c2)(a + b + c)(a^2 + b^2 + c^2).

Thus, we the inequality becomes 

a4a(a2+ab+b2)+b4b(b2+bc+c2)+c4c(a2+ac+c2)a2+b2+c2a+b+c\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 a2+b2+c2a+b+ca+b+c3\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.)

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

1a3(b+c)+1b3(a+c)+1c3(b+a)32\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 a2,b2a^2, b^2 and c2c^2 respectively.

Now, we have by Titu's Lemma,

1a3(b+c)+1b3(a+c)+1c3(b+a)=1a2a(b+c)+1b2b(a+c)+1c2c(b+a)\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)}

(1a+1b+1c)22(ab+bc+ca)=(ab+bc+ca)22(ab+bc+ca)\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)}

ab+bc+ca23(abc)232=32\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

2.2. Prove that for all positive real numbers a,b,c,da, b, c, d

ab+2c+3d+bc+2d+3a+cd+2a+3b+da+2b+3c23\frac{a}{b + 2c + 3d} + \frac{b}{c + 2d + 3a} + \frac{c}{d + 2a + 3b} + \frac{d}{a + 2b + 3c} \ge \frac{2}{3}

3.3. Problem link

4.4. (IMO 2005) 

Prove that for all positive real numbers x,y,zx, y, z such that xyz1xyz \ge 1, then

x2+y2+z2x5+y2+z2+x2+y2+z2y5+x2+z2+x2+y2+z2z5+y2+x23\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
5 years, 9 months ago

No vote yet
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Will there be a part 3 with ore problems??

Eddie The Head - 5 years, 9 months ago

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

Daniel Chiu - 5 years, 9 months ago

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

Priyansh Sangule - 5 years, 2 months ago

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

Daniel Liu - 5 years, 9 months ago

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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.

Daniel Liu - 5 years, 9 months ago

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Actually, Anqi had made a post on Cauchy-Schwarz.

Zi Song Yeoh - 5 years, 9 months ago

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@Zi Song Yeoh Would you be kind enough to give me the link? Thanks.

Daniel Liu - 5 years, 9 months ago

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@Daniel Liu https://brilliant.org/discussions/thread/elementary-techniques-used-in-the-imo-internatio-V/

Zi Song Yeoh - 5 years, 9 months ago

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Edited. Thanks for pointing out.

Zi Song Yeoh - 5 years, 9 months ago

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Hints on Prob 4??? :(

Eddie The Head - 5 years, 9 months ago

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

Calvin Lin Staff - 5 years, 9 months ago

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Whoa!!A little hint?? :p

Eddie The Head - 5 years, 9 months ago

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@Eddie The Head It involves finding an identity which is almost impossible to find.

ZK LIn - 5 years, 9 months ago

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

Dinesh Chavan - 5 years, 9 months ago

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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

x2+y2+z2x5+y2+z2+x2+y2+z2y5+x2+z2+x2+y2+z2z5+y2+x23\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

(x2+y2+z2)(1x5+y2+z2+1y5+x2+z2+1z5+y2+x2)3(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

1x5+y2+z2+1y5+x2+z2+1z5+y2+x23x2+y2+z2\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 (x5+y2+z2)(yz+y2+z2)(x5yz+y2+z2)2(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 xyz1xyz≥1 we get (x5+y2+z2)(yz+y2+z2)(x5yz+y2+z2)2(x2+y2+z2)2(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 1x5+y2+z2yz+y2+z2(x2+y2+z2)2z2+y22+y2+z2(x2+y2+z2)2\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 yy and zz, we get; 1y5+x2+z2x2+z22+x2+z2(x2+y2+z2)2\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

1z5+y2+x2x2+y22+x2+y2(x2+y2+z2)2\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.

Dinesh Chavan - 5 years, 9 months ago

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

Eddie The Head - 5 years, 9 months ago

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Thanks

Dinesh Chavan - 5 years, 9 months ago

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how can we solve question no 2

Rahul Rawat - 3 years, 9 months ago

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For problem 4, (x5+y2+z2)(1x+y2+z2)(x2+y2+z2)2(x^5+y^2+z^2)\left(\frac1x + y^2 + z^2\right) \geq (x^2 + y^2 + z^2)^2     x2+y2+z2x5+y2+z21x+y2+z2x2+y2+z2yz+y2+z2x2+y2+z2\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, cycx2+y2+z2x5+y2+z2cyc(yz+y2+z2)x2+y2+z2\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} =2(x2+y2+z2)+(xy+yz+zx)x2+y2+z2=2+xy+yz+zxx2+y2+z23= \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, x2+y2+z2xy+yz+zx    12[(xy)2+(yz)2+(zx)2]0.x^2 + y^2 + z^2 \geq xy + yz +zx \iff \frac12[(x-y)^2 + (y-z)^2 + (z-x)^2] \geq 0.

Rishabh Dhiman - 1 year, 5 months ago

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For problem 2, cycab+2c+3d=cyca2ab+2ca+3da(a+b+c+d)2cyc(ab+2ac+3ad)\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)} =(a+b+c+d)24(ab+bc+cd+da+ca+bd)=a2+b2+c2+d24(ab+bc+cd+da+ca+bd)+12= \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 a2+b2+c2+d24(ab+bc+cd+da+ca+bd)16    3(a2+b2+c2+d2)2(ab+bc+cd+da+ca+bd)\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)     (ab)2+(bc)2+(cd)2+(da)2+(bd)2+(ac)20\iff (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + (b-d)^2 + (a-c)^2 \geq 0

Rishabh Dhiman - 1 year, 5 months ago

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