An inequality question. Please help me. :) (1 Viewer)

[~shinigami~]

What is the easiest way to prove:

1. a³ + b³ + c³ ≥ 3abc

2. Given that x,y and z are positive real numbers, x+y+z=1 and (x+y+z)/3 ≥ √xyz.

Prove 1/x³ +1/y³ + 1/z³ ≥ 81
Thank You in advance.

 

[Affinity]

(x+y)/2 >= sqrt(xy)

(x+y+z)/3
= {[(x+y+z)/3 + x]/2 + [y + z]/2}/2
>=[sqrt((x+y+z)x/3) + sqrt(yz)]/2
>=((x+y+z)*(xyz)/3)^(1/4)

so

[(x+y+z)/3]^4 >= ((x+y+z)*(xyz)/3)
[(x+y+z)/3]^3 >= xyz
(x+y+z)/3 >cuberoot(xyz)

now, the first question is obvious. for the 2nd notice that if k<1 cuberoot(k)>sqrt(k).

the last question should also be quite straightforward.

 

[airie]

For the first inequality, should a, b, c be all non-negative numbers? For example, if a = -2, b = -3, c = -4, LHS would be -99, whereas RHS equals -72, so LHS is not greater than or equal to RHS.

Assuming a, b, c are positive, notice that a³ + b³ + c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) = (a+b+c){1/2[(a-b)²+(b-c)²+(c-a)²]}. Since squares are non-negative ie. the second bracket is non-negative, and a+b+c is non-negative (assuming a, b, c are non-negative), a³ + b³ + c³ - 3abc >= 0 ie. a³ + b³ + c³ >= 3abc.

(x+y)/2 >= sqrt(xy)

(x+y+z)/3
= {[(x+y+z)/3 + x]/2 + [y + z]/2}/2
>=[sqrt((x+y+z)x/3) + sqrt(yz)]/2
>=((x+y+z)*(xyz)/3)^(1/4)

so

[(x+y+z)/3]^4 >= ((x+y+z)*(xyz)/3)
[(x+y+z)/3]^3 >= xyz
(x+y+z)/3 >cuberoot(xyz)

I see that you've proven AM-GM for x, y, z, taking x+y+z and yz to be positive, so I take that my assumption of the three numbers being positive is correct

EDIT: For the second result, use the first to get that 1/x³ + 1/y³ + 1/z³ >= 3/xyz. And since x+y+z=1, and by AM-GM (which applies to all positive reals) (x+y+z)/3 >= cuberoot(xyz), one obtains xyz <=1/27 (as both sides of the inequality are positive, so you could just cube both sides). Therefore 1/x³ + 1/y³ + 1/z³ >= 3/xyz >= 3*27 = 81

 

[~shinigami~]

Thank You so much Affinity and Airie.

Now, I'm off to study even more for 4U assessment tomorrow. Wish me luck.

 

[Affinity]

good luck mate

 

[~shinigami~]

good luck mate

Thanks. Uh-oh, here I am procrastinating again.

Hey, affinity, if you don't mind. Could you help me out again. There's only one question in my school's past assessment paper that I couldn't do and it's really bugging me.


It is given that z₁ and z₂ are two complex numbers representing points A and B with
|z₁| = |z₂| = 1 and arg(z₂) = 2arg(z₁) with 0 < arg(z₁) < ^pi_/2.

i) Draw an Argand diagram placing points A and B on it.
ii) Show arg(z₂ + 1) = arg(z₁)
iii) Show |z₂ + 1| = 2arg(z₁)

I know it's not an inequality but I didn't want to make another thread.

Thanks again in advance. ^_^

 

[Affinity]

(O) - (Z1) - (Z1 + 1) - (1) is a rhombus and things should follow

you can also use algebra:
z1 = cos(t) + isin(t)
z2 = cos(2t) + isin(2t)

1 + z2 = cos(2t) + 1 + isin(2t)
= 2cos^2(t) + 2icos(t)sin(t)
= 2cos(t) * [ cos(t) + isin(t)]
so arg(z2 + 1) = t = arg(z1)
and |z2 + 1| = 2cos(t) <- (part iii's answer is wrong I think)

 

[airie]

I think part (iii) is definitely wrong. You can't have a length equal to an angle

To solve the other two parts geometrically, sketch the situation on an argand diagram (points representing z1 and z2 both fall on the unit circle) and use circle geometry. Another hint: z2+1 is just the vector from the point (-1,0) to the point representing z2

 

[jyu]

I think part (iii) is definitely wrong. You can't have a length equal to an angle

Why not?
Try to sketch {z: |z| = arg(z)}.

 

[airie]

Why not?
Try to sketch {z: |z| = arg(z)}.

Right. So you mean that the VALUE of the length units equals that of the angle in rads.

So how to you solve part (iii)? Just wondering.

 

[jyu]

Right. So you mean that the VALUE of the length units equals that of the angle in rads.

So how to you solve part (iii)? Just wondering.

Affinity has shown the solution.
I was pointing out that you have made an incorrect statement to justify that the given solution was wrong.