inequality (1 Viewer)

[Premus]

hey

how do you Show that

x1 + x2 > root of (x1x2)

Thanks!

 

[xiao1985]

hey

how do you Show that

x1 + x2 > root of (x1x2)

Thanks!

the inequality does nto stand...

try
x1 = -1
x2 = -1

x1 + x 2 = - 2

sqrt ( -1 x -1) = 1

-2>1 (?) contradiction... therefore the arguemtn is nto true....

 

[redruM]

it would stand if there was a condition like x₁ and x<sub2</sub> > 0, or < 0.

i dont know...havent done maths in ages.

 

[Premus]

sorry, it says

x1 >1 and x2>1

 

[redruM]

square both sides....and expand LHS

LHS = x₁² + 2 x₁x₂ + x₂²

using convention(?)
x₁² is always +ve
similarly,
x₂² is always +ve

and since x₁ > 1 and x₂ > 1, 2x₁x₂ is always +ve.

.'. LHS > RHS

i know there will be some flaws in my explanation, but i think my idea is correct.

edit: made some adjusments.

 

[xiao1985]

ok, in that case =)

edit: >< a min late =p

 

[Affinity]

since when did you start frequenting this forum xiao?

 

[CM_Tutor]

Shorter: If x₁ > 1 and x₂ > 1, then (&radic;x₁ - &radic;x₂)² > 0
Hence, x₁ + x₂ > 2&radic;(x₁x₂)

Now, 2&radic;(x₁x₂) > &radic;(x₁x₂) as x₁x₂ > 0 [you can square root and then add &radic;(x₁x₂) to prove this, if needed.]

So, x₁ + x₂ > &radic;(x₁x₂), as required.

 

[xiao1985]

since when did you start frequenting this forum xiao?

since the great affinity convinced me so... =D

well, to be honest, since the beauty of mathematics truely touched me =p

ok ok, confession: since i was really bored last nite ><

 

[CrashOveride]

a b and c are any three +ve real numbers then (a+b+c)/3 >= cube root of(abc)
a+b+c = 1, show 1/a + 1/b + 1/c >= 9

 

[CM_Tutor]

a b and c are any three +ve real numbers then (a+b+c)/3 >= cube root of(abc)

One approach:

Let x, y and z be positive reals.

(a) Show that x² + y² &ge; 2xy

(b) Hence, show that x² + y² + z² - xy - yz - zx &ge; 0

(c) By multiplying the expression in (b) by x + y + z, and making a substitution, show that for a, b and c positive reals:

(a + b + c) / 3 &ge; cubert(abc)​

I suggest you KNOW THIS PROOF. The AM-GM inequailty pops up quite often, and it is very useful to be able to write out proofs for n = 2, 3 and 4.

 

[CM_Tutor]

a+b+c = 1, show 1/a + 1/b + 1/c >= 9

First, show that, for reals x and y, (x / y) + (y / x) = (x² + y²) / xy &ge; 2

Then, consider the expansion of (a + b + c) * [(1 / a) + (1 / b) + 1 / c)]