Applications: Named Inequalities

The triangle inequality says that one side of a triangle is never longer than the other two sides together, and is equal to them only when the triangle is flat. Applied to each side in turn, it traps the third side between the difference and the sum of the other two.

1. Call the length of the direct road d km. W, D and S are the corners of a triangle, and the route through the depot is two of its sides, so the triangle inequality gives d ≤ 13 + 8 = 21. A route through the depot is never shorter than the direct road.
2. The side WD is also no longer than the other two sides together: 13 ≤ d + 8. Subtract 8 from both sides to get d ≥ 5.
3. (a) The direct road is at least 5 km and at most 21 km long. It is 21 km only when D lies on the direct road between W and S. It is 5 km only when S lies on the road from W to D, 5 km from W and 8 km from D.
4. The route through the depot is 13 + 8 = 21 km. Being no more than 20% longer than the direct road means 21 ≤ 1.2d. Divide both sides by 1.2: d ≥ 17.5.
5. (b) The rule allows the route through the depot when the direct road is from 17.5 km to 21 km long; 21 km is the longest the direct road can be, by (a). Check: when d = 17.5, the route is 21 ÷ 17.5 = 1.2 times the direct road, exactly 20% longer.

answer(a) Least 5 km, when S lies on the road from W to D; greatest 21 km, when D lies on the direct road between W and S; (b) direct roads from 17.5 km to 21 km long

techniqueThe Triangle Inequality

Three lengths make a triangle when each one is less than the sum of the other two. It is enough to test the longest length: if it is less than the sum of the other two, each shorter length is certainly less than the sum of the rest. A sum that only equals the longest length gives a flat triangle.

1. There are 10 ways to choose three of the five lengths. For each choice, compare the longest rod with the sum of the two shorter rods.
2. The choices 20, 30, 40, 20, 40, 50, 30, 40, 50, 30, 50, 70 and 40, 50, 70 pass, since 50 > 40, 60 > 50, 70 > 50, 80 > 70 and 90 > 70. The choices 20, 30, 50, 20, 50, 70 and 30, 40, 70 give sums equal to the longest rod, and 20, 30, 70 and 20, 40, 70 give sums of 50 and 60, short of 70.
3. (a) Five of the ten choices make a frame. The choices 20, 30, 50, 20, 50, 70 and 30, 40, 70 fail by equality: the two shorter rods lie flat along the longest, so they weld into a straight bar, not a triangle.
4. For (b), call the third rod c cm. If the 70 cm rod is the longest, it must be shorter than the other two together: 70 < 30 + c, so c > 40. If the third rod is the longest, c < 30 + 70 = 100.
5. (b) So 40 < c < 100, and in whole centimeters the shortest third rod is 41 cm and the longest is 99 cm. Check: 30 + 41 = 71, which is more than 70, and 30 + 70 = 100, which is more than 99.

answer(a) 5 of the 10 choices; 20, 30, 50, 20, 50, 70 and 30, 40, 70 fail by equality and would weld into a straight bar; (b) shortest 41 cm, longest 99 cm

techniqueThe Triangle Inequality

The shoulder, the elbow and the tip are the corners of a triangle, which is flat in the extreme positions. The triangle inequality bounds the distance from the shoulder to the tip above by the sum of the links and below by the longest link minus the others.

1. Let the tip be r cm from the shoulder. The triangle with corners at the shoulder, the elbow and the tip has sides 50, 30 and r, so the triangle inequality gives r ≤ 50 + 30 = 80 and 50 ≤ 30 + r, which is r ≥ 20.
2. (a) The least distance is 20 cm, with the second link folded back along the first, and the greatest is 80 cm, with the arm straight. Turning the elbow moves the tip smoothly between the two, so every distance from 20 cm to 80 cm can be reached.
3. With a third link of c cm, the greatest reach is 50 + 30 + c = 80 + c cm, with all three links in a line. To reach 90 cm, 80 + c ≥ 90, so c ≥ 10.
4. For the least reach, the first link is no longer than the way round from the elbow along the other two links to the tip and straight back to the shoulder: 50 ≤ 30 + c + r, so r ≥ 20 − c, with equality when both links fold back. To come within 5 cm of the shoulder, 20 − c ≤ 5, so c ≥ 15.
5. (b) Both conditions hold when c ≥ 15, so the shortest third link is 15 cm. Check: with links of 50, 30 and 15 cm, the tip reaches from 50 − 30 − 15 = 5 cm out to 50 + 30 + 15 = 95 cm, which covers every distance from 5 cm to 90 cm.

answer(a) Least 20 cm, with the second link folded back along the first; greatest 80 cm, with the arm straight; (b) 15 cm

techniqueThe Triangle Inequality

The AM–GM inequality says that for positive numbers p and q, p + q ≥ 2√pq, with equality only when p = q. When a total is a sum of two terms whose product is fixed, the right-hand side is a fixed number, so it is the least value of the total, reached when the two terms are equal.

1. The area gives xy = 200. The fencing is the two ends and the front, 2x + y meters.
2. The two terms 2x and y have a fixed product: 2x × y = 2 × 200 = 400. By the AM–GM inequality, 2x + y ≥ 2√400 = 2 × 20 = 40.
3. (a) Equality holds when the two terms are equal, 2x = y. Then x × 2x = 200, so x² = 100, x = 10 and y = 20. The least fencing is 40 m, with ends of 10 m and a front of 20 m.
4. The cost is 10 × 2x + 40y = 20x + 40y dollars. The product 20x × 40y = 800xy = 800 × 200 = 160000 is fixed, so 20x + 40y ≥ 2√160000 = 2 × 400 = 800.
5. (b) Equality holds when 20x = 40y, so each term is half of 800: 20x = 400 gives x = 20, and 40y = 400 gives y = 10. The least cost is $800, with ends of 20 m and a front of 10 m. Check: 20 × 10 = 200 square meters, and 20 × 20 + 40 × 10 = 400 + 400 = 800.

answer(a) 40 m of fencing, with x = 10 m and y = 20 m; (b) $800, with x = 20 m and y = 10 m

techniqueThe Arithmetic Mean–Geometric Mean Inequality

An average speed is the total distance divided by the total time, not the mean of the speeds. The AM–GM inequality, squared, compares the two, because the average speed over equal distances has the product ab on top.

1. The ride out takes 30 ÷ 20 = 1.5 hours and the ride back takes 30 ÷ 30 = 1 hour, so the 60 km take 2.5 hours. The average speed is 60 ÷ 2.5 = 24 km/h.
2. In general, a distance s each way takes sa + sb = s(a + b)ab hours. Dividing the total distance 2s by this time gives the average speed 2aba + b.
3. By the AM–GM inequality, a + b2 ≥ √ab. Both sides are positive, so squaring gives (a + b)²4 ≥ ab, which is (a + b)² ≥ 4ab. Divide both sides by 2(a + b): a + b2 ≥ 2aba + b.
4. (a) Her average speed is 24 km/h, less than the mean of the speeds, 25 km/h. The two are equal only when a = b, the equality case of AM–GM, so whenever the speeds out and back differ, the average speed is less than their mean.
5. For (b), an average of 20 km/h over 60 km means 60 ÷ 20 = 3 hours in all. The ride out at 15 km/h takes 30 ÷ 15 = 2 hours, which leaves 1 hour for the 30 km back.
6. (b) She must ride back at 30 ÷ 1 = 30 km/h. Check: 2 × 15 × 3015 + 30 = 90045 = 20. The mean of 15 and 30 is 22.5, more than 20, as the inequality in (a) says it must be.

answer(a) 24 km/h; the average speed is at most a + b2, with equality only when a = b; (b) 30 km/h

techniqueThe Arithmetic Mean–Geometric Mean Inequality

Driving faster saves hours of wages but burns fuel faster. The wage term falls as the speed rises and the fuel term rises, and their product does not depend on the speed, so AM–GM gives the least total. When the speed that gives equality is not allowed, the gap from the least total, written as a square, shows which allowed speed comes closest.

1. The run takes 400v hours, and each hour costs 36 + v²100 dollars. So the cost is 400v × 36 + 400v × v²100 = 14400v + 4v dollars.
2. The product of the two terms is fixed: 14400v × 4v = 57600. By the AM–GM inequality, 14400v + 4v ≥ 2√57600 = 2 × 240 = 480.
3. (a) Equality needs 14400v = 4v, so v² = 3600 and v = 60. The least cost is $480, at 60 km/h. Check: at 60 km/h the wages are 1440060 = $240 and the fuel is 4 × 60 = $240.
4. The limit rules out 60 km/h. The gap from the least cost is a square divided by v: 14400v + 4v − 480 = 4(v − 60)²v. Below 60 km/h, a lower speed makes (v − 60)² larger and v smaller, so the gap grows, and the cheapest allowed speed is the limit, 50 km/h.
5. (b) At 50 km/h the cost is 1440050 + 4 × 50 = 288 + 200 = $488, which is $8 more than in (a). Check: the gap is 4 × (50 − 60)²50 = 40050 = 8.

answer(a) $480, at 60 km/h; (b) $488, which is $8 more

techniqueThe Arithmetic Mean–Geometric Mean Inequality

The Cauchy–Schwarz inequality says that a dot product is never more than the product of the lengths of the two vectors, u · v ≤ |u| |v|, with equality only when the vectors are parallel. Here the length of one vector is fixed by the weights and the length of the other by the budget.

1. Let the spends on search, social media and radio be x, y and z dollars, with x + y + z = 4900. The new customers number 2√x + 3√y + 6√z, the dot product of (2, 3, 6) with (√x, √y, √z).
2. The first vector has length √2² + 3² + 6² = √49 = 7, and the second has length √x + y + z = √4900 = 70. By Cauchy–Schwarz the dot product is at most 7 × 70 = 490.
3. Equality needs the vectors parallel: (√x, √y, √z) = k(2, 3, 6) for a number k. Then x + y + z = 4k² + 9k² + 36k² = 49k² = 4900, so k² = 100, k = 10, and the square roots of the spends are 20, 30 and 60.
4. (a) The most is 490 new customers, from $400 on search, $900 on social media and $3600 on radio. Check: 2 × 20 + 3 × 30 + 6 × 60 = 40 + 90 + 360 = 490, and 400 + 900 + 3600 = 4900.
5. For a budget of B dollars the same argument gives at most 7√B new customers, reached by a split in the same proportions. To bring 700, the budget needs 7√B ≥ 700, so √B ≥ 100.
6. (b) The least budget is $10000. With any smaller budget, even the best split brings fewer than 7 × 100 = 700 new customers.

answer(a) 490 new customers, from $400 on search, $900 on social media and $3600 on radio; (b) $10000

techniqueThe Cauchy–Schwarz Inequality

The Cauchy–Schwarz inequality, u · v ≤ |u| |v|, gives a lower bound on a sum of squares when a plain sum is fixed. Split each current into a part whose square is the heat loss and a part that depends only on the resistance, so that the dot product is the total current.

1. Let the currents in the three cables be x, y and z amps, with x + y + z = 30. The total heat loss is P = 0.2x² + 0.3y² + 0.6z² watts.
2. Take u = (√0.2x, √0.3y, √0.6z) and v = (1√0.2, 1√0.3, 1√0.6). Their dot product is x + y + z = 30. The length of u is √P, and the length of v squared is 10.2 + 10.3 + 10.6 = 5 + 103 + 53 = 10.
3. By Cauchy–Schwarz, 30 ≤ √P × √10. Square both sides: 900 ≤ 10P, so P ≥ 90 watts.
4. Equality needs u parallel to v, so √RI = k√R in every cable, which is I = kR for one number k. Then 5k + 103k + 53k = 10k = 30, so k = 3.
5. (a) The least loss is 90 W, with 3 ÷ 0.2 = 15 A, 3 ÷ 0.3 = 10 A and 3 ÷ 0.6 = 5 A in the three cables. Each cable then has the same voltage drop, RI = 3 volts, which is how current divides between cables joined side by side. Check: 0.2 × 225 + 0.3 × 100 + 0.6 × 25 = 45 + 30 + 15 = 90.
6. (b) With two cables, the length of v squared is 5 + 103 = 253, so 900 ≤ 253P and P ≥ 108 W, with 18 A and 12 A. The least loss rises by 108 − 90 = 18 W. Check: 0.2 × 324 + 0.3 × 144 = 64.8 + 43.2 = 108.

answer(a) 90 W, with 15 A, 10 A and 5 A in the 0.2, 0.3 and 0.6 ohm cables; (b) it rises by 18 W, to 108 W

techniqueThe Cauchy–Schwarz Inequality

A logarithm grows more slowly than any power of n, and an exponential grows faster than any power. So a count of nlog₂ n steps, whatever number multiplies it, falls below n² once n is large enough, and each extra reading doubles the work of the group search.

1. B takes fewer steps than A when n² < 32nlog₂ n. Divide both sides by n: n < 32log₂ n, which compares n with a multiple of its logarithm.
2. Try powers of 2, where log₂ n is a whole number. At n = 128: 32 × 7 = 224, more than 128. At n = 256: 32 × 8 = 256, equal. At n = 512: 32 × 9 = 288, less than 512.
3. (a) The ratio nlog₂ n is 2 at n = 2 and rises for every n ≥ 3, so it reaches 32 only once, at n = 256. B takes fewer steps for 2 ≤ n < 256, the two take the same number, 65536, at n = 256, and A takes fewer from then on.
4. One second is 1000000 steps. For B, n² ≤ 1000000 gives n ≤ 1000. For the group search, 2ⁿ ≤ 1000000 gives n ≤ log₂ 1000000 ≈ 19.9, so n ≤ 19: 2¹⁹ = 524288 and 2²⁰ = 1048576.
5. A chip 64 times as fast carries out 64000000 steps in a second. For B, n² ≤ 64000000 gives n ≤ 8000. For the group search, 64 = 2⁶ adds 6 to the logarithm: n ≤ 19.9 + 6 = 25.9, so n ≤ 25, since 2²⁵ = 33554432 and 2²⁶ = 67108864.
6. (b) In one second B sorts up to 1000 readings and the group search handles up to 19. On the faster chip these become 8000 and 25: eight times as many readings for B, but only 6 more for the group search.

answer(a) B takes fewer steps for 2 ≤ n < 256, the same number at n = 256, and more for n > 256; (b) B sorts up to 1000 readings and the group search handles up to 19; on the faster chip, 8000 and 25

techniquePolynomial, Exponential and Logarithmic Growth

The computers' speed is an exponential in the years, while the cost of a finer grid is only a power of the refinement. Setting one equal to the other and writing both as powers of 2 turns the question into an equation between indices.

1. The speed doubles once every 2 years, so after y years the computers are 2^y/2 times as fast. A model k times finer runs in 1 hour when 2^y/2 ≥ k⁴.
2. A spacing of 5 km is k = 2 times finer, so the run takes 2⁴ = 16 times as long. The computers must be 16 = 2⁴ times as fast: 2^y/2 ≥ 2⁴, so y2 ≥ 4.
3. (a) The 5 km model first runs in 1 hour after 8 years. Check: 8 years hold 4 doublings, and 2⁴ = 16.
4. After 24 years there are 12 doublings, so the computers are 2¹² = 4096 times as fast. The model needs k⁴ ≤ 4096, and 4096 = 2¹² = 8⁴, so k ≤ 8. The finest spacing is 10 ÷ 8 = 1.25 km.
5. The rival model needs k⁶ ≤ 4096, and 4096 = 2¹² = 4⁶, so k ≤ 4. Its finest spacing is 10 ÷ 4 = 2.5 km.
6. (b) After 24 years the model runs at a spacing of 1.25 km and the rival at 2.5 km. Both keep improving: k⁴ = 2^y/2 gives k = 2^y/8, and k⁶ = 2^y/2 gives k = 2^y/12, so a higher power only slows the exponential growth in k and never stops it.

answer(a) 8 years; (b) 1.25 km for the model, and 2.5 km for the rival

techniquePolynomial, Exponential and Logarithmic Growth