(Y+Z-X)(Z+X-Y)(X+Y-Z) is less than or equal to XYZ

(X,Y,Z> 0)

Have you heard of the AM-GM (Arithmetic Mean - Geometric Mean) inequality? It says that for n positive numbers:

x₁ , x₂ , ... x_n

the following is true:

(x₁ x₂ x₃ ...x_n )^(1/n) < = (x₁ +x₂ +x₃ + ... + x_n )/n

with equality only when x₁ = x₂ = ... = x_n . Perhaps you can do it using this...

Michael

Well, to tackle this question, fist note that there can be at most one term on the LHS is negative, in which case the inequality is immediate since LHS < 0 < RHS.

So we can suppose X,Y,Z are sides of a triangle.
Let a=Y+Z-X, b=Z+X-Y and c=X+Y-Z

Thus, AM-GM on pairs (a,b), (b,c) and (c,a) gives
(as a,b,c > = 0)

(a+b)/2 > = (ab)^1/2
(b+c)/2 > = (bc)^1/2
(c+a)/2 > = (ca)^1/2

Multiplying the three inequalities together, and with the help of a+b=2Z etc. gives

XYZ > = abc

which is the required inequality after subsituting back a=Y+Z-X etc.

Kerwin.

Kerwin - that is quite amazing. Incidentally I've just had a look in one of my exercise books, and noticed that I did this question myself a while ago. But it took me ages to realise you had to apply AM-GM to pairs of (X+Y-Z), (X-Y+Z), (-X+Y+Z). (Actually I think I applied it to a² b, a² c, b² a, b² c, c² a, c² b, abc and abc in your notation so did it in one stage). But how did you work out what to do so quickly?

Michael

I see your logic, but how do you know that a,b,c> 0? Perhaps I do not totally understand as I have not seen the AM-GM inequality before. How do you know that a=b=c? If the part saying that this equation is only true when x₁ =x₂ =x_n is not needed then there are no problems, but if so I am not sure that I entirely understand.

Thanks,

Brad

Brad - Suppose, without loss of generality, that
c=X+Y-Z < 0,
then we have: Z> X+Y
so we have: X+Z> 2X+Y> Y and Y+Z> X+2Y> X
so: a,b> 0.

This is true because of the condition X,Y,Z> 0. Geometrically speaking, the values a/2,b/2,c/2 is the distance of the vertices of a triangle with lengths X,Y,Z from the incircle, so it must be positive if X,Y,Z> 0 forms a triangle.(the case a,b or c = 0 is trivial.)

In the AM-GM inequality, equality occurs if and only if all the x's are equal. There are various proofs of AM-GM, by induction, geometry or variational methods. Perhaps you should try to find a proof. (N.B. simple minded approach for induction does NOT work)
[See here for a proof. - The Editor]

Michael - Have you taken a look at 1964 IMO question 2? If a,b,c are sides of a triangle then prove
a² (b+c-a)+b² (c+a-b)+c² (a+b-c)< = 3abc
I will not spoil you pleasure for proving this here, but the first step of one approach is by the above inequality.

Kerwin

I'll have a think about the question but don't expect anything immediate.

Thanks,

Michael

Right - got it. Let x,y,z > 0. Apply AM-GM to:

xy² , xz² , x² y, yz² , x² z and y² z

6xyz < = xy² + xz² + x² y + yz² + x² z + y² z

Add the contents of the right hand side twice to both sides. Also add 6xyz to both sides and re-arrange:

2(xy² + 2xyz + xz² + yx² + 2xyz + yz² + zx² + 2xyz + zy² ) < = 3(x² y + x² z + xzy + xz² + y² x + xyz + y² z + yz² )

Factorise:

2[x(y+z)² + y(x+z)² + z(x+y)² ] < = 3(x+y)(x+z)(y+z)

Let c = 1/2(x+y), b = 1/2(x+z), a = 1/2(y+z).

And the answer comes out.

However equality can only occur if x = y = z. (Provided two of x,y,z can't be zero, which I don't think they can because this would make a side of the triangle zero. Well maybe that's allowed, I don't know.) Otherwise I don't think equality can occur.

Yours,

Michael

By the way, do you know if the continuous form of AM-GM is ever useful? I assume this should be:

where f(x), g(x) > 0 in the range of the integral, all integrals wrt x and between the same limits. With equality only when f(x) is constant. g(x) is the weighting function.

I can't really think this is going to be as useful as the discrete AM-GM.

Yours,

Michael

Michael,

One of the approach for the IMO inequality is to rearrange
(b+c-a)(c+a-b)(a+b-c)
to give : a² (b+c-a)+b² (c+a-b)+c² (a+b-c)-2abc,
from which the inequality is immediate.

Another approach is to rearrange the LHS to give
a(b² +c² -a² )+b(c² +a² -b² )+c(a² +b² -c² )
and using the cosine rule, we have
2abc(cos A+cos B+cos C), and the maximum of
cos A+cos B+cos C
is 3/2 (by variation, for example). The way you did it was also fine.

Anyway, I am not sure about the application of continuous form of AM-GM. Perhaps asking an expert would help.

Kerwin

Oh, that's interesting. I never knew that cos A + cos B + cos C was limited by 3/2. Easy to prove this by differentiation. It looks like sin A + sin B + sin C has an upper limit of 2, but obviously tan A + tan B + tan C can't have an upper limit.

Anyway I like the way geometry can be introduced into the problem via the cosine rule. I guess AM-GM has many geometrical interpretations. For example you could say the n = 2 case is telling you that if you fix its area the least possible perimeter of a rectangle is when its a square, which is easy to prove geometrically as well.

Thanks,

Michael

There was another part to the question. If z and also z1, z2, z3.. zn are complex numbers with w = (1/n) x (z1 + z2 + z3 + ... zn) then

sum(mod(z-zi)² ) > = n mod(z-w)²

Well as it happens this is considerably easier than the preceeding part.

z simply causes an origin shift so we can set z = 0 without loss of generality.

Next write z1 = a1 + i b1, z2 = a2 + i b2, etc. Expand out both sides of the inequality, substituting everything in. And you are left needing to prove that:

n x (a1² + a2² + ... + an² ) + n*(b1² + b2² + ... + bn² ) > = (a1 + a2 + ... + an)² + (b1 + b2 + ... + bn)²

where a1 a2 ... and b1 b2 ... are all real.

To show this we first show that:

n*(a1² + a2² + ... + an² ) > = (a1 + a2 + ... + an)²

If you expand out the right hand side and re-arrange you can see that this inequality is equivalent to:

(a1-a2)² + (a1-a3)² + ... + (a1-an)² + (a2-a3)² + (a2-a4)² + ... + (a2-an)² + (a3-a4)² + ... etc > = 0

Basically on the left hand side the square of the difference of every pair of a1 a2 a3 ... an appears exactly once. Exactly the same is true about b1 b2 ... bn. The truth of these inequalities is evident because all squares are non-negative. Therefore the inequality to be proven holds.

Yours,

Michael

Michael - Have you heard of Cauchy-Schwartz inequality? It states that

equality iff (if and only if) there exist non-zero l, m such that

la_i=mb_i for all i.
Thus the inequality you want to show can be done in one line with all the a's(or b's)=1.

Oh, that's good. I had heard of Cauchy-Schwarz recently from elsewhere on NRICH, but I don't know its formal derivation. I guess it can be interpreted geometrically as: the square of the cosine of the angle between two n-D vectors is less than or equal to 1, and only equal to 1 if the two vectors are parallel. Which is good. I'll have a think about formal justification; please don't tell me yet!

Actually having thought about it, the inequality is simply a re-arrangement of:

OK, that's fair enough. If we specify they're non-zero then it is okay to just use one constant.

I see what you mean about defining angles in higher dimensions. Instead of plugging in the cosine rule (or the dot product) formula, I thought you may be able to do the following:

Let a and b be the vectors in n-D space you want to find the angle between. Find any set of constants p,q,r,s such that:

(pa + qb ) and (ra + sb ) are perpendicular and unit. Now simply find the angle between:

ri - pj and -si + qj

where i and j are the standard perpendicular vectors. However we're still going to need the dot product to show the above vectors are perpendicular so I guess there isn't much point in this approach, you might as well just plug in with the dot product as you say. You could check with Pythagoras I guess... Anyway, it really doesn't matter.

Yours,

Michael