Inequalities - Mixing Variable Technique
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Inequalities: Inequalities: Mixing Variab Variables les by: Adrian Adrian Tang Email: Email: tang @ math.ucalg math.ucalgary ary.ca .ca I will assume that prior to reading these notes that you know the QM-AM-GM-HM Inequality, Jensen’s Inequality, Inequality, Cauchy-Schw Cauchy-Schwarz arz Inequality, Inequality, Rearrangement Rearrangement Inequality, Inequality, Chebyshev Inequality, Inequality, Holder Inequality, Inequality, Muirhead Inequality and Schur’s Inequality very well. Not to mention substitution techniques such as Ravi Substitution and Trignometric Substitution. Warm-Up Problems
≥ 0 with x + y + z = 1. Prove that 0 ≤ xy + yz + zx − 2xyz ≤ 277 . 2. Let x,y,z ≥ 0 with x + y + z = 3. Prove that x5 y5 z5 3 ≥ + + . y 3 + z 3 z 3 + x3 x3 + y 3 2 1. Let x,y,z
3. Let x, y,z > 0 such that
1 1 1 + + = 2. 1+x 1+y 1+z
Prove that
4. Let x,y,z
1 1 1 + + 1 + 4x 4 x 1 + 4y 4 y 1 + 4z 4z
≥ 0 and n ∈ N. Prove that a + 2b2b a +b +c ≥ 3
n
n
n
n
≥ 1. n
+
n
b + 2c2c c + 2a2a 3
+
3
.
5. Let x, y,z > 0 with x + y + z = 3xyz 3 xyz.. Prove that 1 1 1 + 3+ 3 3 x y z
≥ 3.
≥ 0 such that x2 + y2 + z2 + xyz = 4. Prove that x + y + z ≤ 3. 7. Let x,y,z ≥ 0 with xy + yz + zx = 2. Prove that 7(x 7(x + y + z )3 − 9(x 9(x3 + y 3 + z 3 ) ≤ 108 108.. 8. Let a,b,c ≥ 0. Prove that a2 + b2 + c2 + 2abc 2abc + 1 ≥ 2(ab 2(ab + bc + ca). ca). 6. Let x,y,z
9. Let ABC be a triangle with circumradius R and inradius r. Prove Prove that that R equality holds.
≥ 2r .
Determ Determine ine when when
10. Let ABC be a triangle with side lengths a,b,c, a,b,c, circumcentre O, centroid G and circumradius R. Prove that 1 2 OG 2 R2 a + b2 + c2 . 9
| | ≤ −
P . Let D,E,F be the feet Erdos-Mordell Erdos-Mordell Inequality: Inequality: Let ABC be a triangle with an interior point P . of the perpendicular from P onto BC,CA,AB , respectively. Then
|P A| + |P B| + |P C | ≥ 2(|P D| + |P E | + |P F |). 1
Inequalities - Mixing Variable Technique
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Mixing Variable Technique What do the following inequalities inequalities have in comm common? on?
∈ R with x2 + y2 + z2 = 9. Prove that 2(x 2( x + y + z ) − xyz ≤ 10. 2. Let x,y,z ≥ 0 such that xy + yz + zx = 1. Prove that 1 1 1 9 ≥ + + . 2 2 2 (x + y ) (y + z ) (z + x) 4 1. Let x,y,z
3. Let x,y,z
≥ 0. Prove that √xx+ y + √yy+ z + √zz+ x ≤ 54 √x + y + z.
All of these inequalities inequalities are symmetric symmetric or cyclic, but have have non-trivia non-triviall equality equality cases. Can you find out what the equality cases are? Equality cases come in the following categories for three-variable inequalities. 1. All three variabl variables es are equal. 2. Exactly Exactly two of the variables ariables are equal. 3. At least one variable variable is the b oundary oundary of the domain of the variables. variables. For example, example, if x,y,z at least one of x,y,z is equal to 0 in the equality case.
≥ 0, then
We will first handle inequalities inequalities that satisfies satisfies one of the first two cases. cases. (i.e. at least two variables variables are equal in the equality equality case). case). Let’s revisit a very very simple inequality inequality..
≥ 0. Prove that x + y + z ≥ 3 √xyz. xyz . √xyz . Let Let f ( f (x,y,z) x,y,z) = x + y + z − 3 xyz. Let t = + Prove that that f ( f (x,y,z) x,y,z ) ≥ f (t,t,z) t,t,z ) ≥ 0. 2 . Prove √
Problem 1: Let x,y,z Proof:
prove this also for t =
3
x y
3
xy. xy .
You can can
How does this help us to solve inequalities inequalities with non-trivia non-triviall equality equality cases? Suppose equality equality holds for x = y . Then when we set t to be equal to one of the means of x and y . Then prove f ( f (x,y,z) x,y,z ) f (t,t,z) t,t,z ) 0. The whole equality case will then depend on the steps in the proof of f ( f (t,t,z) t,t,z ) 0. Of course, we want to check the equality cases first prior to attempting to solve the problem.
≥
≥
≥
Let’s dissect an equality problem. Problem 2: Let a,b,c > 0. Prove that
1 1 1 + + 2a2 + bc 2b2 + ca 2c2 + ab
≥ (a + b8+ c)2 .
Proof: We conjecture that (a,b,c ( a,b,c)) = (t,t, 0) for some t > 0 and its permutations is when the equality case
holds. holds. Note Note that a = b = c does not give equality equality. Of course, there is likely a way to solve this problem using Schur’s Schur’s Inequality Inequality.. But I am presenting presenting this proof as a means means of presentin presentingg a technique technique,, and will not argue one way or another that this method is cleaner than the other methods.
2
Inequalities - Mixing Variable Technique Let f ( f (x,y,z) x,y,z) =
2x2
3
1 1 1 + 2 + 2 + yz 2y + zx 2z + xy
− (x + y8+ z)2 .
Again, we conjecture that our equality case occurs at ( a,b,c) a,b,c) = (t,t, 0). i.e. i.e. a = b and c = 0. Hence, Hence, let’s let’s assume that c = min a,b,c . Most likely, this fact will become important to solve this problem.
{
}
b Let t = a+ prove that that f ( f (a,b,c) a,b,c) f ( f (t,t,c) t,t,c) with equality if and only if a = b = t and 2 . We will need to prove f ( f (t,t,c) t,t,c) 0 with equality if and only if c = 0. Use the fact that a c, b c to prove this and establish the equality case for good.
≥
≥
≥
≥
Other minimum minimum and maximum maximum cases can be done using elementary elementary differential differential calculus. We present present the MV techniqu techniquee for another inequality inequality problem. problem. Howeve However, r, this problem contains contains an initial initial condition, condition, which influences are choice of t (the mean of two of the variables) when applying the MV method. Problem 3: Let x,y,z be real numbers such that x2 + y 2 + z 2 = 9. Prove that
2(x 2(x + y + z )
− xyz ≤ 10 10..
Playing around with the equality case will yield that (x,y,z ( x,y,z)) = (2, (2, 2, 1) and its permutations are equal-
−
2
2
y ity cases for this inequality inequality. This problem also allows x,y,z to be negative. negative. Let t = x + 2 . Why did we choose this t? This is because (x,y,z ( x,y,z)) = (t,t,z) t,t,z ) needs to satisfy x2 + y 2 + z 2 = 9 for the MV method to work.
Let z = min x,y,z . We may want to try to prove that f ( f (x,y,z) x,y,z) by basic algebraic manipulation,
{
}
≤ f ( f (t,t,z) t,t,z) and f ( f (t,t,z) t,t,z ) ≤ 10. Note that
− f ( f (t,t,z) t,t,z ) = 2(x 2(x + y − 2t) − z (xy − t2 ). By QM − AM inequality, we have x + y − 2t ≤ 0. But xy ≤ t2 . However, However, f ( f (x,y,z) x,y,z) ≤ f ( f (t,t,z) t,t,z ) is therefore true if z ≤ 0. This is a good observation since if z ≥ 0, then x, y ≥ z ≥ 0 implying x,y,z ≥ 0. This case will f (x,y,z) x,y,z)
be left for the reader.
Now suppose z 0. Then Then f ( f (x,y,z) x,y,z) algebraic manipulation,
≤
≤ f ( f (t,t,z). t,t,z ).
We will proceed proceed now to prove prove that f ( f (t,t,z) t,t,z )
f ( f (t,t,z) t,t,z) = 4t + 2z 2z Since 2t 2t2 + z 2 = 9, let t2 =
(9 − z )/2. Then we have 9 − z 9−z 2
2
≤ 10. 10.
By
− t2z.
2
3
z − 5z · z = 2 18 − 2z 2 + . 2 2 2 We want to prove that the maximum of g (z) is 10 in the range −3 ≤ z ≤ 0. Hence, −4z + 3z2 − 5 . g (z ) = √ 2 18 − 2z 2 g(z ) = f ( f (t,t,z) t,t,z ) = 4
+ 2z 2z
−
Before you think that solving for g (z ) = 0 is a lot of work, remember that we ”know” that z = 1 is a solution. solution. (Remember (Remember the conjectured conjectured equality equality case.) This will help in the factoring factoring when we solve for g (z ). We will leave to the reader to prove that in the domain of z [ 3, 0] that z = 1 is the only solution to g (z ) = 0. We will then see that g ( 3) < g ( 1) and g (0) < g ( 1). This implies z = 1 is the maximum of
−
−
∈− −
−
3
−
−
Inequalities - Mixing Variable Technique
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g in the domain z [ 3, 0]. We certainly hope at this point that g ( 1) = 10. And yes this is true. Hence, for z 0, f ( f (x,y,z) x,y,z) f ( f (t,t,z) t,t,z ) 10.
∈− ≤
≤
−
≤
The only remaining case is x,y,z
≥ 0. We will leave for the reader to complete this case.
We will present one more example. Problem 4: Let a,b,c
≥ 0 such that ab + bc + ca = 1. Prove that √ √ ≥
1 1 1 + + 2 2 ( a + b) (b + c) (c + a)2
√
≥ 94 .
Note that (x,y,z ( x,y,z)) = (1/ (1/ 3, 1/ 3, 1/ 3) and (1, (1 , 1, 0) are both equality cases. So we need to choose t such that f (a,b,c) a,b,c) f (t,t,c) t,t,c) 9/4. But what what is t this time so that ab + bc + ca = 1 is satisfied? satisfied? Well, clearly, clearly, 2 we want a = b = t to work. Hence, we want t R such that t + 2tc 2tc = 1.
≥
∈
Now, prove that f ( f (a,b,c) a,b,c) f ( f (t,t,c) t,t,c) 9/4. A few tips to make the following to make things better. These tips work for any variables such that ab + bc + ca = 1 = 2t 2t2 + ct. ct.
≥
≥
• t2 + 2ct 2ct = 1 = ab + bc + ca ⇔ (t + c)2 = (a √ + c)(b )(b + c√ ) √ √ • a + b − 2t = (a ( a + c) + (b ( b + c) − 2(c 2(c + t) = ( a + c − b + c)2 = (a ( a − b)2 /( a + c + b + c). Then
f ( f (a,b,c) a,b,c)
− f ( f (t,t,c) t,t,c)
=
1 (a+b)2
− 41
=
4t2 −(a+b)2 4(a+b)2 t2
=
(2t−a−b)(2t+a+b) 4(a+b)2 t2
+
t2
+
1 (b+c)2
+
1 (a+c)2
− ( + )(2 + ) a c b c
(a−b)2 (a+c)2 (b+c)2
+
(a−b)2 (a+c)2 (b+c)2
− b)2( 4(√ + −+(2√ ++ +) () + ) + ( + ) 1( + ) ). Now, use we assume that a ≥ c, b ≥ c (and therefore t ≥ c) to finish the proof. If an initial condition is say x + y + z = 3, then a way to prove that f ( f (x,y,z) x,y,z) ≤ f ( f (t,t,z) t,t,z ) (or ≥) where = (a
a c
t a b b c 2 a b
2 2
t
a c
2
b c
2
t = (x + y)/2 is to fix z , and prove using calculus that f ( f (x,y,z) x,y,z) is maximized (or minimized) when x = y .
Exercises 1. Let a,b,c
≥ 0 such that a + b + c = 1. Prove that (c − a)2 + c + (a(a − b)2 ≥ √3. a + (b (b − c)2 + b + (c 2. Let a,b,c ≥ 0 such that ab + bc + ca = 1. Prove that 1 1 1 5 ≥ + + . a+b b+c c+a 2 3. Let A,B,C be A,B,C be angles of a non-obtuse triangle. Prove that sin A + sin B + sin C cos A + cos B + cos C
4
≤ 1+
√2 2
.
