STEP II 2007 question 7 solution

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STEP II 2007 question 7 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 7 solution

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 , 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 , 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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