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# 1. Let x, y, and z be real numbers such that Find the maximum value of x + y + z.

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1. Let x, y, and z be real numbers such that $$x^2 + y^2 + z^2 = 1.$$ Find the maximum value of x + y + z.

2. Let a, b, c, d be positive real numbers such that a + b + c + d = 1. Find the minimum value of $$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}.$$

3. The sequence $$(a_n)$$ is defined recursively by $$a_0=1, a_1=\sqrt[19]{2},$$ and $$a_n=a_{n-1}a_{n-2}^2$$ for $$n\geq 2.$$ What is the smallest positive integer k such that the product $$a_1a_2\cdots a_k$$ is an integer?

Jul 12, 2019

#1
+23846
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Let $$x$$, $$y$$, and $$z$$ be real numbers such that $$x^2 + y^2 + z^2 = 1$$.

Find the maximum value of $$x + y + z$$.

The Cauchy–Schwarz inequality states that for all vectors $${\displaystyle u}$$ and  $${\displaystyle v}$$ of an inner product space it is true that
$${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle ,}$$

where $${\displaystyle \langle \cdot ,\cdot \rangle }$$ is the inner product.

$$\text{Let \vec{u} = \begin{pmatrix} x \\y\\z \end{pmatrix} } \\ \text{Let \vec{v} = \begin{pmatrix} 1 \\1\\1 \end{pmatrix} }$$

$$\begin{array}{|rcll|} \hline \langle \mathbf {u} ,\mathbf {v} \rangle &=& \begin{pmatrix} x \\y\\z \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& x+y+z \\\\ \langle \mathbf {u} ,\mathbf {u} \rangle &=& \begin{pmatrix} x \\y\\z \end{pmatrix}\begin{pmatrix} x \\y\\z \end{pmatrix} \\ &=& x^2+y^2+z^2 \\\\ \langle \mathbf {v} ,\mathbf {v} \rangle &=& \begin{pmatrix} 1 \\1\\1 \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& 1^2+1^2+1^2 \\ &=& 3 \\\\ \hline |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2} &\leq& \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle \\ (x+y+z)^2 &\le& (x^2+y^2+z^2)\cdot 3 \\ x+y+z &\le& \sqrt{x^2+y^2+z^2}\cdot \sqrt{3} \quad | \quad x^2+y^2+z^2 = 1 \\ x+y+z &\le& 1\cdot \sqrt{3} \\ \mathbf{ x+y+z } & \mathbf{\le} & \mathbf{\sqrt{3}} \\ \hline \end{array}$$

The maximum value of $$x + y + z$$ is $$\mathbf{\sqrt{3}}$$

Jul 12, 2019
#4
+106533
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Thanks, heureka.....that's a pretty neat method for solving this one......

CPhill  Jul 12, 2019
#6
+23846
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Thank you CPhill !

heureka  Jul 14, 2019
#2
+23846
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3.
The sequence $$(a_n)$$ is defined recursively by $$a_0=1,\ a_1=\sqrt[19]{2}$$, and $$a_n=a_{n-1}a_{n-2}^2$$ for $$n\geq 2$$.
What is the smallest positive integer k such that the product $$a_1a_2\cdots a_k$$ is an integer?

$$\begin{array}{|l|r|rcl|l|} \hline & k & \text{product } a_1a_2\cdots a_k \\ \hline a_2 = a_1a_0^2=a_1^1 & 2 & a_1a_2 &=& a_1^2 & =2^{\frac{2}{19}} & = 1.07569058622 \\ \hline a_3 = a_2a_1^2=a_1^1a_1^2=a_1^3 & 3 & a_1a_2a_3 &=& a_1^5 & =2^{\frac{5}{19}} & = 1.20010271958 \\ \hline a_4 = a_3a_2^2=a_1^3a_1^2=a_1^5& 4 & a_1a_2\cdots a_4&=& a_1^{10} & =2^{\frac{10}{19}} & = 1.49375896165 \\ \hline a_{5} = a_4a_3^2=a_1^{5}a_1^{6}=a_1^{11} & 5 & a_1a_2\cdots a_5 &=& a_1^{21} & =2^{\frac{21}{19}} & = 2.15138117244 \\ \hline a_{6} = a_5a_4^2=a_1^{11}a_1^{10}=a_1^{21} & 6 & a_1a_2\cdots a_6 &=& a_1^{42} & =2^{\frac{42}{19}} & = 4.62844094913 \\ \hline a_{7} = a_6a_5^2=a_1^{21}a_1^{33}=a_1^{43} & 7 & a_1a_2\cdots a_7 &=& a_1^{85} & =2^{\frac{85}{19}} & = 22.2184182818 \\ \hline a_{8} = a_7a_6^2=a_1^{43}a_1^{42}=a_1^{85} & 8 & a_1a_2\cdots a_8 &=& a_1^{170} & =2^{\frac{170}{19}} & = 493.658110947 \\ \hline a_{9} = a_8a_7^2=a_1^{85}a_1^{86}=a_1^{171} & 9 & a_1a_2\cdots a_9 &=& a_1^{341} & =2^{\frac{341}{19}} & = 252752.952805 \\ \hline a_{10} = a_9a_8^2=a_1^{171}a_1^{170}=a_1^{341} & 10 & a_1a_2\cdots a_{10} &=& a_1^{682} & =2^{\frac{682}{19}} & = 2^{35.8947368421} \\ \hline a_{11} = a_{10}a_{9}^2=a_1^{341}a_1^{342}=a_1^{683} & 11 & a_1a_2\cdots a_{11} &=& a_1^{1365} & =2^{\frac{1365}{19}} & = 2^{71.8421052632} \\ \hline a_{12} = a_{11}a_{10}^2=a_1^{683}a_1^{682}=a_1^{1365} & 12 & a_1a_2\cdots a_{12} &=& a_1^{2730} & =2^{\frac{2730}{19}} & = 2^{143.684210526} \\ \hline a_{13} = a_{12}a_{11}^2=a_1^{1365}a_1^{1366}=a_1^{2731} & 13 & a_1a_2\cdots a_{13} &=& a_1^{5461} & =2^{\frac{5461}{19}} & = 2^{287.421052632} \\ \hline a_{14} = a_{13}a_{12}^2=a_1^{2731}a_1^{2730}=a_1^{5461} & 14 & a_1a_2\cdots a_{14} &=& a_1^{10922} & =2^{\frac{10922}{19}} & = 2^{574.842105263} \\ \hline a_{15} = a_{14}a_{13}^2=a_1^{5461}a_1^{5462}=a_1^{10923} & 15 & a_1a_2\cdots a_{15} &=& a_1^{21845} & =2^{\frac{21845}{19}} & = 2^{1,149.73684211} \\ \hline a_{16} = a_{15}a_{14}^2=a_1^{10923}a_1^{10922}=a_1^{21845} & 16 & a_1a_2\cdots a_{16} &=& a_1^{43690} & =2^{\frac{43690}{19}} & = 2^{2,299.47368421} \\ \hline a_{17} = a_{16}a_{15}^2=a_1^{21845}a_1^{21846}=a_1^{43691} & \mathbf{17} & a_1a_2\cdots a_{17} &=& a_1^{87381} & =2^{\frac{87381}{19}} & \mathbf{= 2^{4599}}\\ & & && & & \text{the product a_1a_2\cdots a_{17}} \\ & & && & & \text{with k=17 is an integer} \\ \hline \end{array}$$

The smallest positive integer $$k$$ such that the product $$a_1a_2\cdots a_k$$ is an integer is $$\mathbf{17}$$

Jul 12, 2019
edited by heureka  Jul 12, 2019
#3
+1

$$\prod_{i=1}^{k} a_{i}= {a_1}^{\frac{{2}^{k+1}-1+\frac{{(-1)}^{k+1}-1}{2}}{3}}={2}^{\frac{{2}^{k+1}-1+\frac{{(-1)}^{k+1}-1}{2}}{57}}$$

Guest Jul 12, 2019
#5
+28357
+3

Here, symmetry is used to tackle questions 1 and 2:

1. Both the constraint and the function to be maximised are symmetrical in x, y and z, so we should expect them all to have the same value at the maximum point. Hence let y=x and z=x so the constraint becomes: 3x2 = 1 That is x = 1/√3 so x+y+z = 3/√3 = √3

2. Similarly, symmetry suggests a and b should take the same value.   Each term in the function to be minimised is of the form k/x where k is a constant.  The rate of change of a term like this with respect to x is -k/x2.  We'd like the rate of change of each term in the function to contribute equally to the minimum value, so we want  -1/a2 = -4/c2  and -1/a2 = -16/d2 so that c = 2a and d = 4a (we already have b = a ) Putting these values in the constraint equation we get  a + a + 2a + 4a = 1  or a = 1/8 Hence b = 1/8,  c = 2/8 and d = 4/8, so 1/a + 1/b + 4/c + 16/d = 64 the minimum value.

Jul 12, 2019