B-Real – Wikipedia


B-Real – Wikipedia

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

B-REAL скачать все песни в хорошем качестве (320Kbps)

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

B-Real – Fire (Dj Nut VIP Remix), Marchesini and Farina vs Max B – Majestade Real (DJ Ilia Kashtan & DJ. и другие музыкальные треки в хорошем качестве (128-320kbps). Dj Nut – B-Real – Fire (Dj Nut VIP Remix)
DJ Emon – B- Real Ft. Damian Marley – Fire (DJ Emon Remix)
DJ B-Real – It’s Easy (Feat. Mouse On Tha Track) [Prod. by DJ B-Real & Doe-Doe]
моэстро – happy b akira kamigawa dj niki – the real busters vol. супер клубняк 2010 евгения волконская 9. лёш
Dj 5150 – 14 Kevin Gates – Goin’ Supa [Prod. By Dj B-Real]
DJ B-Real – You Don’t Know About (Feat. Doodie Da Juvie and 1st Class Carson) [Prod. by DJ B-Real and J Buc
2po2 – Prishtinali Ft. DJ Blunt, Lumi B, Lyrical Son, Capital T, Real 1, Mixey & DJ Flow
DJ Quik – Real Women (Feat. Jon B) (Prod. by DJ Quik)
DJ Murres – Puppet Master (feat. Dr. Dre & B-Real) prod. by DJ Murres
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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