Question

The arithmetic-geometric mean (AM-GM) inequality: if a,b ≥ 0, then
`(a+b)/(2) \geq √(ab)`. (1)

(a) Let a, b, c, d ≥ 0 and consider the positive real numbers
`√(ab)` and
`√(cd)`. Use (1) to prove the four-variable AM-GM inequality:


`(a+b+c+d)/(4) \geq \sqrt[4]{abcd}`. (2)

(b) Let a, b, c ≥ 0 and define
`m= (a+b+c)/(3)`.

(i) Show that
`m=(a+b+c+m)/(4)`.

(ii) Hence use (2) to prove the three-variable AM-GM inequality:


`(a+b+c)/(3) \geq \sqrt[3]{abc}`.

Answer 1

(a) The inequality follows from the binomial theorem.

`(a - b)^2 = a^2 - 2ab + b^2 \ge 0 \implies a^2 + 2ab + b^2 = (a+b)^2 \ge 4ab \\\\ \implies \frac{(a+b)^2}4 \ge ab \\\\ \implies \frac{a+b}2 \ge √(ab) ~~~~~~~~ (1)`

By the same reasoning,

`\frac{c+d}2 \ge √(cd)`

Adding these results together, we have

`\frac{a+b}2 + \frac{c+d}2 = \frac{a+b+c+d}2 \ge √(ab) + √(cd)`

and since
`√(ab)` and
`√(cd)` are both positive, they also satisfy the AM-GM inequality,

`\frac{√(ab) + √(cd)}2 \ge \sqrt{√(ab)*√(cd)} = \sqrt[4]{abcd}`

and the 4-variable result follows,

`\frac{a+b+c+d}2 \ge 2*\frac{√(ab)+√(cd)}2 \ge 2 \sqrt[4]{abcd} \\\\ \implies \frac{a+b+c+d}4 \ge \sqrt[4]{abcd} ~~~~~~~~ (2)`

(b.i) With
`m=\frac{a+b+c}3`, we have

`\frac{3m}4 = \frac{a+b+c}4 \implies m - \frac m4 = \frac{a+b+c}4 \\\\ \implies m = \frac{a+b+c+m}4`

(b.ii) From (2) it follows that

`\frac{a+b+c+m}4 \ge \sqrt[4]{abcm} = \sqrt[4]{abc*\frac{a+b+c+m}4}`

Using the result from (b.i), this is equivalent to

`\frac{a+b+c}3 \ge \sqrt[4]{abc*\frac{a+b+c}3}`

Take 4th powers on both sides.

`\left(\frac{a+b+c}3\right)^4 \ge abc * \frac{a+b+c}3`

Divide both sides by
`\frac{a+b+c}3`.

`\left(\frac{a+b+c}3\right)^3 \ge abc`

Finally, take the cube root of both sides.

`\frac{a+b+c}3 \ge \sqrt[3]{abc}`

as required.