One idea, one pictureDrag θ,φ to find the maximum of E=xy+2yz. Of the two chains of bounds on the right, which one can get a ✓ on every link?
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In one sentence
AM–GM and the Cauchy inequality are the same kind of tool: an inequality sign plus an equality condition. In a solution you often string them into a chain
E≤E1≤E2≤⋯≤M.
E≤M is certainly true; but whether M is the maximum depends on whether there is one point where every link is an equality at the same time. So the test for choosing tools is not “which one applies” but “which combination has equality conditions that are compatible”.
Example: x² + y² + z² = 1; find the maximum of E = xy + 2yz
Step 1 / 5
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Two chains
Left chain: write E as y(x+2z), apply Cauchy to x+2z, then AM–GM to the product that is left; this gives E≤25. Right chain: AM–GM twice, giving E≤23. Both are true inequalities.
E≤25≈1.118E≤23
Try “all three equal” first
The right chain can never be completed
Left chain: the Cauchy link
The AM–GM link
Stop 1 · Fractions: AM–GM twice vs Cauchy once
a+b=1, a,b>0. Find the minimum of a1+b8.
Predict first
“a1+b8≥2ab8≥82≈11.31.” Is the minimum 82?
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.
No. The equality conditions of the two steps are b=8a and a=b, which cannot both hold, so 82 is never reached, and a lower bound that is never reached means the true minimum is larger. Cauchy gets there in one step: the minimum is (1+22)2=9+42≈14.66. Drag a below and watch the left chain get a ✓ on every link at a≈0.26.
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Stop 2 · Symmetric expressions: AM–GM twice can be fine
Clashing equality conditions are the mistake; “using it twice” is not a mistake in itself.
One problem, several solutions
Let a,b,c>0. Prove that (a+b+c)(a1+b1+c1)≥9.
AM–GM twice
a+b+c≥33abc and a1+b1+c1≥3abc3; multiplying gives ≥9. The equality condition of both steps is a=b=c, so they can hold at the same time, and 9 is a minimum that is actually reached.
CostTwo steps, but both equality conditions are a = b = c, so they are compatible.
Cauchy once
Take (a,b,c) and (a1,b1,c1): every product is 1, so the left side is ≥(1+1+1)2=9. Equality needs all the ratios a:a1 to be equal, i.e. a=b=c.
CostOne step, and again the equality condition is a = b = c.
Stop 3 · Radicals: the geometric face of Cauchy
x+y=4. Find the minimum of x2+1+y2+4.
The two terms are the lengths of the vectors (x,1) and (y,2). The Cauchy inequality implies the Minkowski inequality (the triangle inequality): ∣u∣+∣v∣≥∣u+v∣, so
x2+1+y2+4≥(x+y)2+(1+2)2=16+9=5.
Equality needs (x,1) and (y,2) to point the same way: y=2x; with x+y=4 this gives x=34, y=38, and then 35+310=5. If you applied AM–GM to each square root separately here (for example x2+1≥2x), equality would need x=1 and y=2, contradicting x+y=4: another chain that can never be completed.
Rule of thumb for choosing tools: first think about where equality should be (you can often guess it), then pick a chain in which every link becomes an equality at that point.
Practice
Exercise 1Basic
a,b>0, a+b=1. Find the minimum of a1+b4.
Hint · direction
AM–GM twice gives 8. Check its two equality conditions.
Hint · method
Use the fraction form of Cauchy: a12+b22≥a+b(1+2)2.
Hint · first step
The right side =9. Equality needs a1=b2, i.e. b=2a.
Full solution
At a=31, b=32 we get 3+6=9. The minimum is 9. (AM–GM twice: ≥2ab4≥8, but equality needs b=4a and a=b, a contradiction, so 8 is never reached.)
Exercise 2Basic
x2+y2+z2=1. Find the maximum of 3xy+4yz.
Hint · direction
Copy the example at the start of this stop: first take out y.
Hint · method
3xy+4yz=y(3x+4z)≤∣y∣⋅5x2+z2 (Cauchy); then apply AM–GM to ∣y∣x2+z2.
Hint · first step
∣y∣x2+z2≤2y2+x2+z2=21, so 3xy+4yz≤25.
Full solution
Two equality conditions: (x,z) points the same way as (3,4) (with the same sign as y), and y2=x2+z2=21. Take y=21 and (x,z)=521(3,4); substituting gives y(3x+4z)=21⋅5225=25. The maximum is 25.
Exercise 3Advanced
Real numbers x,y satisfy x+y=4. Find the minimum of x2+1+y2+4.
Hint · direction
The two square roots are the lengths of two vectors.
Hint · method
∣u∣+∣v∣≥∣u+v∣ with u=(x,1), v=(y,2).
Hint · first step
∣u+v∣=(x+y)2+9=25.
Full solution
So the sum is ≥5. Equality needs u,v to point the same way, y=2x, i.e. x=34, y=38, and then 35+310=5. The minimum is 5.
Exercise 4Challenge
x,y>0, x+y=1. Find the minimum of (x+x1)(y+y1).
Hint · direction
“x+x1≥2 and y+y1≥2, so it is ≥4”: check the equality conditions.
Hint · method
Both of those equalities need x=y=1, which contradicts x+y=1. Try another approach: expand, and write everything in terms of t=xy.
Hint · first step
Expanding gives xy+xy1+xyx2+y2, and x2+y2=1−2t, so the expression is t+t2−2, where 0<t≤41.
Full solution
h(t)=t+t2 is decreasing on (0,2) (AM–GM would give h≥22, but equality needs t=2, which again is out of reach), so the minimum is at t=41, i.e. x=y=21: 41+8−2=425. Check: (21+2)2=425. The minimum is 425.
Using AM–GM twice with clashing equality conditions
✗
The wrong step: a+b=1, a,b>0. a1+b8≥2ab8≥232=82, so the minimum is 82.
What went wrong? Think first, then open
DiagnosisEquality in the first step needs a1=b8, i.e. b=8a; equality in the second step, ab≤41, needs a=b. The two conditions cannot hold together, so 82≈11.3 is never reached. The true minimum is about 14.66.
✓
The right step: Use Cauchy (fraction form), which gets there in one step: a12+b(22)2≥a+b(1+22)2=9+42, with equality when b=22a.
“When you chain several inequalities, their equality conditions must hold together”
AM–GM and Cauchy Together ← → step · F full screen · Esc exit