Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality

Continuing from: Part 1

Part 2


Example 1: (Iran 1998) Suppose that x,y,z1 x,y,z \geq 1 and 1x+1y+1z=2 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2 . Prove that:

x+y+zx1+y1+z1 \sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} .

Solution: The condition given is considerably weird. In a sense, we would like to involve the variables more than just on the denominators. We have 2 2 approaches, either use a substitution or try to rewrite it in a nicer form. It seems that the latter works. We notice that most conditions have the constant given as 1 1 so it is pretty motivated to write the hypothesis as:

x1x+y1y+z1z=1 \frac{x-1}{x} + \frac{y-1}{y} + \frac{z-1}{z} = 1 .

Now notice how "magically" we have (x1),(y1), (x-1), (y-1), \ldots on the numerator. In fact, due to the fact that we can square both sides, we can immediately see how to apply Cauchy:

cycx=(cycx)(cycx1x)(cycx1)2 \sum_{cyc}x = ( \sum_{cyc}x)( \sum_{cyc} \frac{x-1}{x}) \geq ( \sum_{cyc} \sqrt{x-1} )^2 . ■

Example 2: (Japan 2004) Let a,b,c a,b,c be positive real numbers with sum 1 1 . Prove that

1+a1a+1+b1b+1+c1c2ab+2bc+2ca \frac{1+a}{1-a} + \frac{1+b}{1-b} + \frac{1+c}{1-c} \leq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a} .

Solution: With a little experience in inequalities, we see some use for a+b+c=1 a + b + c = 1 :

  1. Substitute into the question

  2. Substitue as a=xx+y+z,b=yx+y+z,etc a = \frac{x}{x+y+z}, b = \frac{y}{x+y+z}, etc

Now, also note that we do NOT want to make the inequality in question more complicated, so we opt to using the former. Also with some experience we see that we do not like fractions such as 1+a1a \frac{1+a}{1-a} or 1a1+a \frac{1-a}{1+a} for the matter. We like to only have the variable and no constants (or vice versa) on the denominator so that we can easily form squares. Therefore, by substituting and dividing both sides by 2 2 to make the RHS ab, \frac{a}{b} , \ldots, we rewrite the inequality as:

32+ab+c+ba+b+ca+bab+bc+ca \frac{3}{2} + \frac{a}{b+c} + \frac{b}{a+b} + \frac{c}{a+b} \leq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}

We isolate the constant and rearrange to get:

cycacb(b+c)32 \sum_{cyc} \frac{ac}{b(b+c)} \geq \frac{3}{2}

Intuitively, we multiply both numerator and denominator of the LHS by ac ac (taken cyclically), and directly apply Cauchy Schwarz,

cycacb(b+c)=cyca2c2abc(a+c)(ab+bc+ca)22abc(a+b+c)32 \sum_{cyc} \frac{ac}{b(b+c)} = \sum_{cyc} \frac{a^2c^2}{abc(a+c)} \geq \frac{(ab+bc+ca)^2}{2abc(a+b+c)} \geq \frac{3}{2}

Example 3: (IMO SL 2011) Given arbitrary real numbers x1,x2,,xn x_1, x_2, \ldots, x_n , prove that:

x11+x12+x21+x12+x22++xn1+x12+x22++xn2 \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} + \ldots + \frac{x_n}{1+x_1^2 + x_2^2 + \ldots + x_n^2} .

Solution: We recall the extremely useful variant of Cauchy:

n(a12+a22++an2)(a1+a2++an)2 n(a_1^2 + a_2^2 + \ldots + a_n^2) \geq (a_1 + a_2 + \ldots + a_n)^2

So, what we do is that, we apply the above inequality directly to the given inequality in question. The motivation behind this step is to generate an n n term. So,

(x11+x12+x21+x12+x22++xn1+x12+x22++xn2)2[(x11+x12)2++(xn1+x12+x22++xn2)2]×n (\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \ldots + \frac{x_n}{1+x_1^2 + x_2^2 + \ldots + x_n^2})^2 \leq [(\frac{x_1}{1+x_1^2})^2 + \ldots + (\frac{x_n}{1+x_1^2+x_2^2 + \ldots + x_n^2})^2] \times n .

Notice also that for k2 k \geq 2 , we have:

(xk1+x12+x22++xk2)2=xk21+x12+x22++xk2)2 (\frac{x_k}{1+x_1^2+x_2^2 + \ldots + x_k^2})^2 = \frac{x_k^2}{1+x_1^2+x_2^2 + \ldots + x_k^2)^2}

xk2(1+x12+x22++xk2)(1+x12+x22++xk12) \leq \frac{x_k^2}{(1+x_1^2+x_2^2 + \ldots + x_k^2)(1+x_1^2+x_2^2 + \ldots + x_{k-1}^2)}

11+x12+x22++xk1211+x12+x22++xk2 \leq \frac{1}{1+x_1^2+x_2^2 + \ldots + x_{k-1}^2} - \frac{1}{1+x_1^2+x_2^2 + \ldots + x_k^2} .

The idea here is that in order to generate n n we might as well try to eliminate the variables by telescoping. Observe finally, that for k=1 k = 1 , we have:

(x11+x12)2x121+x12=111+x12 (\frac{x_1}{1+x_1^2})^2 \leq \frac{x_1^2}{1+x_1^2} = 1 - \frac{1}{1+x_1^2}

I leave my reader to finish the argument.■

These daily instalments will be paused until Friday afternoon, Singapore time because I am going overseas.

Note by Anqi Li
6 years, 6 months ago

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Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page?

Calvin Lin Staff - 5 years, 8 months ago

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can you give me refence for this material ??

link or other ?

Takwa Tri Subekti Subekti - 6 years, 6 months ago

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@Anqi Li

Mardokay Mosazghi - 6 years, 1 month ago

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Nice my friend.

Toan Pham Quang - 6 years, 6 months ago

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Dũng Phạm Văn - 6 years, 6 months ago

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Awesome Post!!!Really helpful....keep them coming!!!!

Eddie The Head - 6 years, 6 months ago

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Great post! and the main thing is you are involving motivation to every problem..Good job Anqi!

Kishan k - 6 years, 6 months ago

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