with equality if and only if y1x1=y2x2=⋯=ynxn. It is just the Cauchy inequality for the two lists yixi and yi: their products are xi, and their squares are yixi2 and yi.
Predict first
The cards 112,122,232 add up to 9.5. What is the value after merging them all into one card?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
Merge everything: the bases of the numerators add to 1+2+3=6, the denominators to 1+1+2=4, giving 462=9.
Only 0.5 is lost in total, and all of it in the first merge: 11 and 12 have different ratios. The second merge combines 23 with 23: the same ratio, so nothing is lost.
Why: merging two cards
Merging can only make it smaller
Step 1 / 6
Loading figure…
Start with two cards
Look only at the first two cards, y1x12 and y2x22. We want to show that merging them into y1+y2(x1+x2)2 does not increase the value.
y1x12+y2x22≥y1+y2(x1+x2)2?
Subtract directly
The loss depends on the ratios
Merge one card at a time
When nothing is lost at all
Higher exponents: the power-sum inequality
A family of relatives
Write the same idea in different ways and you get several standard variants. Each one comes from choosing the two lists well:
Fraction form (the one above): ∑yixi2≥∑yi(∑xi)2, yi>0.
When the numerator is not a square: for ai,bi>0, write biai as aibiai2 to get ∑biai≥∑aibi(∑ai)2, with equality if and only if all the bi are equal.
Hölder's inequality: for p,q>1 with p1+q1=1, ∑aibi≤(∑aip)1/p(∑biq)1/q (ai,bi>0). The case p=q=2 is Cauchy.
The power-sum inequality: the last step above, with exponents m+1 and m.
Practice
Exercise 1Basic
Positive numbers x,y,z satisfy x+y+z=2. Find the minimum of x1+y1+z4.
Hint · direction
All three numerators are squares: 1=12, 4=22. And the sum of the denominators is known.
Hint · method
Use the fraction form: x12+y12+z22≥x+y+z(1+1+2)2.
Hint · first step
The right side is 216=8. You still have to check that equality can actually occur.
Full solution
Equality needs x1=y1=z2, i.e. x:y:z=1:1:2; with x+y+z=2 this gives x=y=21, z=1, and then 2+2+4=8. The minimum is 8.
Exercise 2Basic
Real numbers x,y,z satisfy x+y+z=6. Find the minimum of x2+2y2+3z2.
Hint · direction
Read x2 as 1x2. All the denominators are positive constants.
Hint · method
1x2+2y2+3z2≥1+2+3(x+y+z)2. (Here x,y,z may be negative: the fraction form only needs positive denominators.)
Hint · first step
The right side =636=6.
Full solution
Equality needs 1x=2y=3z; with x+y+z=6 this gives (x,y,z)=(1,2,3), and substituting gives 1+2+3=6. The minimum is 6.
Exercise 3Advanced
Positive numbers x,y satisfy x+y=1. Find the minimum of x21+y28.
Hint · direction
The denominators are squares, and the numerators 1 and 8 are both cubes.
Hint · method
Use the power-sum inequality with m=2: x213+y223≥(x+y)2(1+2)3.
Hint · first step
The right side is 127=27. Equality needs x1=y2.
Full solution
From y=2x and x+y=1 we get x=31, y=32, and then 9+4/98=9+18=27. The minimum is 27.
Exercise 4Challenge
Let a,b,c>0. Prove that b+ca2+c+ab2+a+bc2≥2a+b+c.
Hint · direction
The numerators are already squares and the denominators are all positive. What is the denominator after merging?
Hint · method
Use the fraction form directly: merge all three cards at once.
Hint · first step
The left side is ≥(b+c)+(c+a)+(a+b)(a+b+c)2.
Full solution
The denominator is 2(a+b+c), so the left side is ≥2(a+b+c)(a+b+c)2=2a+b+c. Equality needs b+ca=c+ab=a+bc; if a>b, then a has the bigger numerator and the smaller denominator, so a bigger ratio. Hence only a=b=c works.
DiagnosisThe left side is 1−2=−1, far below 8. The fraction form comes from the Cauchy inequality applied to (yixi) and (yi), and yi needs yi>0.
✓
The right step: Before using ∑yixi2≥∑yi(∑xi)2, make sure every denominator is positive.
“Only positive denominators can be merged”
Exhibit
Merging when the numerator is not a square
✗
The wrong step: 11+13≥1+1(1+3)2=8.
What went wrong? Think first, then open
DiagnosisThe left side is only 4. The merging formula needs each numerator to be a square: yx2. When the numerator is 3, the matching x is 3, not 3.
✓
The right step: Write 112+1(3)2≥2(1+3)2≈3.73; that is correct. Or first multiply top and bottom by the same number to make the numerator a square.
“Read each numerator as a square first, then add up the things being squared”
Variants and Generalizations of Cauchy ← → step · F full screen · Esc exit