# B-Real – Wikipedia

B-Real, is an American rapper. Since 1991, he has been one of two lead rappers in the hi.

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B-Real:] Straight up we set the fire we blaze ’em and do it properly Sonically all we need is the honest and the Cali weed Roll up the sensei when you .

### Feliz Cumpleaños Louis Freese A.K.A B-Real de Cypress Hill 🎂 🎈 🎉 🎶 @breal Cypres

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2po2 – Prishtinali Ft. DJ Blunt, Lumi B, Lyrical Son, Capital T, Real 1, Mixey & DJ Flow

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DJ B-Real – Let Me Do Me [Prod. by DJ B-Real]

#### Cypress Hill Hip hop music, Cypress hill, Hip hop and r&b

a+b+c)(1/a+1/b+1/c)>(1+1+1)2=9. Share. Share a link to this answer. Given that $a, b, c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$

This is how I tried doing it:

Let $p= a + b + c, $ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.

Using AM>GM for $p, q$, I get:

$$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2. $
$$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9, $$
which is the opposite of what had to be proven. What did I do wrong?
asked Jan 29, 2015 at 10:54
You are using $x+1/x\ge 2$ in the opposite direction. You can do it by AM-GM as follows:
$$a+b+c>3\sqrt[3]{abc}\ \ \land\ \ 1/a+1/b+1/c>3/\sqrt[3]{abc}\ \ \implies\ \ (a+b+c)(1/a+1/b+1/c)>9$$

Alternatively, you can use the Cauchy-Schwarz inequality:

answered Jan 29, 2015 at 11:00

user2345215user234521516. 2k2 gold badges28 silver badges57 bronze badges

$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ight) = 3+\sum_{cyc}\left(\frac{b}{a}+\frac{a}{b}\ight) \geq 3+6 = \color{red}{9}$$

answered Jan 29, 2015 at 11:06

Jack D’AurizioJack D’Aurizio348k41 gold badges374 silver badges813 bronze badges

Well, although everyone has answered it, but I think $A. M \geq H. M. $ should also do the work. It’s more better in this case. Moreover no one has addressed as to why the OP is wrong.

We know that $\frac{a+b+c}{3} \geq \frac{3}{\frac1a + \frac1b + \frac1c}$.

Henceforth, we get $(a+b+c)(\frac1a + \frac1b + \frac1c) \geq 9$.

Coming to the error in your attempt, in the step where you have written $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
Clearly, $x+\frac1x \geq 2$, which gives the RHS of the above inequality greater than 3. So therefore, you have said that if $A>B$ and $A>C$ then $B>C$, which is of course not true. Here we cannot compare $B$ and $C$. Hence we couldn’t proceed by your method.

answered Jan 29, 2015 at 14:35

Rohinb97Rohinb971, 6322 gold badges15 silver badges31 bronze badges

without a loss of generality, we can also say that

So, putting it all together we have

answered Jan 29, 2015 at 13:52

John JoyJohn Joy7, 5491 gold badge23 silver badges30 bronze badges

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