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The AM–GM inequality

Exercise 11 gave a geometric proof that the arithmetic mean of two positive numbers (a) và (b) is greater than or equal khổng lồ their geometric mean. We can also prove this algebraically, as follows.

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Since (a) and (b) are positive, we can define (x = sqrta) & (y = sqrtb). Then

eginalign*(x-y)^2 geq 0 quad&impliesquad x^2+y^2-2xy geq 0 \ &impliesquad dfracx^2+y^22 geq xyendalign*

and so

This is called the AM–GM inequality. Cảnh báo that we have equality if and only if (a = b).


Find the range of the function (f(x) = x^2 + dfrac1x^2), for (x eq 0).


Using the AM–GM inequality,

< f(x) = x^2 + dfrac1x^2 geq 2sqrtx^2 imes dfrac1x^2 = 2. >

So the range of (f) is contained in the interval (<2,infty)). Chú ý that (f(1) = 2) và that (f(x) o infty) as (x o infty). Since (f) is continuous, it follows that, for each (y geq 2), there exists (x geq 1) with (f(x) = y). Hence, the range of (f) is the interval (<2,infty)).

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Exercise 15

Find the arithmetic, geometric và harmonic means of (3,4,5) và write them in ascending order. Prove that the harmonic mean of two positive real numbers (a) và (b) is less than or equal to lớn their geometric mean.

The AM–GM inequality can be generalised as follows. If (a_1,a_2,dots,a_n) are (n) positive real numbers, then

a_1a_2dots a_n.>

The next exercise provides a proof of this result.

Exercise 16

Find the maximum value of the function (f(x) = log_e x - x), for (x > 0). Hence, deduce that (log_e x leq x-1), for all (x > 0). Let (a_1,a_2,dots,a_n) be positive real numbers & define (A = dfraca_1 + a_2 + dots + a_nn). By successively substituting (x = dfraca_iA), for (i=1,2,dots,n), into the inequality from part (a) và summing, show that < log_eBigl(dfraca_1a_2dots a_nA^nBigr) leq 0. > By exponentiating both sides of the inequality from part (b), derive the generalised AM–GM inequality.


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