Continuing from: Part 1

**Part 2**

**Applications**:

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

$\sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}$.

*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 $2$ 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 $1$ so it is pretty motivated to write the hypothesis as:

$\frac{x-1}{x} + \frac{y-1}{y} + \frac{z-1}{z} = 1$.

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

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

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

$\frac{1+a}{1-a} + \frac{1+b}{1-b} + \frac{1+c}{1-c} \leq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a}$.

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

Substitute into the question

Substitue as $a = \frac{x}{x+y+z}, b = \frac{y}{x+y+z}, etc$

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 $\frac{1+a}{1-a}$ or $\frac{1-a}{1+a}$ 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 $2$ to make the RHS $\frac{a}{b} , \ldots$, we rewrite the inequality as:

$\frac{3}{2} + \frac{a}{b+c} + \frac{b}{a+b} + \frac{c}{a+b} \leq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$

We isolate the constant and rearrange to get:

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

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

$\sum_{cyc} \frac{ac}{b(b+c)} = \sum_{cyc} \frac{a^2c^2}{abc(a+c)} \geq \frac{(ab+bc+ca)^2}{2abc(a+b+c)} \geq \frac{3}{2}$■

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

$\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} + \ldots + \frac{x_n}{1+x_1^2 + x_2^2 + \ldots + x_n^2}$.

*Solution*: We recall the extremely useful variant of Cauchy:

$n(a_1^2 + a_2^2 + \ldots + a_n^2) \geq (a_1 + a_2 + \ldots + a_n)^2$

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 $n$ term. So,

$(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \ldots + \frac{x_n}{1+x_1^2 + x_2^2 + \ldots + x_n^2})^2 \leq [(\frac{x_1}{1+x_1^2})^2 + \ldots + (\frac{x_n}{1+x_1^2+x_2^2 + \ldots + x_n^2})^2] \times n$.

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

$(\frac{x_k}{1+x_1^2+x_2^2 + \ldots + x_k^2})^2 = \frac{x_k^2}{1+x_1^2+x_2^2 + \ldots + x_k^2)^2}$

$\leq \frac{x_k^2}{(1+x_1^2+x_2^2 + \ldots + x_k^2)(1+x_1^2+x_2^2 + \ldots + x_{k-1}^2)}$

$\leq \frac{1}{1+x_1^2+x_2^2 + \ldots + x_{k-1}^2} - \frac{1}{1+x_1^2+x_2^2 + \ldots + x_k^2}$.

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

$(\frac{x_1}{1+x_1^2})^2 \leq \frac{x_1^2}{1+x_1^2} = 1 - \frac{1}{1+x_1^2}$

I leave my reader to finish the argument.■

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

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## Comments

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

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can you give me refence for this material ??

link or other ?

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@Anqi Li

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Nice my friend.

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easy

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

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

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