STEP II 2007 question 7 solution - The Student Room
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STEP II 2007 question 7 solution

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Since

\displaystyle \frac{\text{d}^2}{\text{d} x^2} \sin x = -\sin x

, and -\sin x < 0 in 0 < x < \pi, \sin x is concave in this interval.

Similarly,

\displaystyle \frac{\text{d}^2}{\text{d} x^2} \ln x = - \frac{1}{x^2},

which is negative for x > 0, (and indeed negative for  x < 0 as well, the only problem being \ln x is not defined there), and thus \ln x is concave in x > 0.

(i) First we note that A + B + C = \pi since they are angles in a triangle.

Applying Jensen's inequality with n = 3, x_1 = A, x_2 = B, x_3 = 3 and f(x) = \sin(x) yields

\displaystyle \frac{1}{3}\left(\sin A + \sin B + \sin C\right) \leq \sin \frac{A + B + C}{3}.

But since A + B + C = \pi, the RHS equals \sin (\pi / 3) = \sqrt 3 / 2, and thus we have

\displaystyle \frac{1}{3}\left(\sin A + \sin B + \sin C\right) \leq \frac{\sqrt{3}}{2}

from which the desired result immediately follows.

(Note that since A, B, C are angles in a triangle, 0 < A, B, C < \pi and so our usage of Jensen's inequality was valid.)

(ii)

Setting, in Jensen's inequality, x_1 = t_1, x_2 = t_2 \cdots x_n = t_n and f(x) = \ln x, we get

\displaystyle \frac{1}{n}\left(\ln t_1 + \ln t_2 + \cdots + \ln t_n\right) \leq \ln \frac{t_1 + t_2 + \cdots + t_n}{n}

Exponentiating both sides (note that since e^x is an increasing function, this preserves the inequality)

\displaystyle \left(t_1t_2\cdots t_n\right)^{1/n} \leq \frac{t_1 + t_2 + \cdots + \t_n}{n} (*)

as required.

a) Setting, in (*),\;\; n = 4,\;\; t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 16, we get

\displaystyle \sqrt[4]{16x^4y^4z^4} \leq \frac{x^4 + y^4 + z^4 + 16}{4}

from which follows (note that we in this step use x, y, z > 0

\displaystyle 2xyz \leq \frac{x^4 + y^4 + z^4 + 16}{4}

8xyz \leq x^4 + y^4 + z^4 + 16

as required.

(Again, the usage of Jensen's inequality was valid, since x^4, y^4, z^4 > 0 which was a requirement for the function to be concave)

b) Let g(x, y, z) = x^5 + y^5 + z^5 - 5xyz.

Set, in [b](*)[/b], n = 5, t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1, t_5 = 1, and we get

\displaystyle \sqrt[5]{x^5y^5z^5} \leq \frac{x^5 + y^5 + z^5 + 1 + 1}{5}

\displaystyle xyz \leq \frac{x^5 + y^5 + z^5 + 2}{5}

\displaystyle 5xyz \leq x^5 + y^5 + z^5 + 2

\displaystyle x^5 + y^5 + z^5 - 5xyz \geq -2

(This inequality is valid whenever x, y, z are positive, because then x^5, y^5, z^5 are positive as well, which was a requirement for [b](*)[/b] to be valid)

Now, this means

g(x, y, z) \geq -2.

But g(x, y, z) attains this value at (x, y, z) = (1, 1, 1), which means that this is the function's minimum. (By definition, since we have just proven that for all (x, y, z), where x, y, z > 0, we have g(1, 1, 1) \geq g(x, y, z))

and so the desired minimum value for the expression is -2.

Solution by ukgea.

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