# Two algebra questions

I have two questions that I need help with:

1) How many single digit even natural number solutions are there for the equation $A+B+C+D = 24$ such that $A+B > C+D$

A)20 B)11 C)16 D)24

2) Three positive real numbers $x,y,z$ are such that $x+y+Z = 1$. which of the following inequalities best discribe the relation between $XY,YZ,ZX$.

A) $xy+yz+zx > 1/3$

B) $xy+yz+zx <1/3$

C) $xy+yz+zx \le 1/3$

D) $xy+yz+zx \le 2/3$

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For the first, do you count $0$ as a natural? Some do, some don't. For the second, please make sure the variables are consistently capitals or lower case. – Ross Millikan Jun 25 '12 at 18:36

For the first question, you have on the one hand that $(A+B)+(C+D)=24$ and on the other hand that $A+B>C+D$. What are the possibilities for $A+B$ and $C+D$? Clearly you must have $A+B>12$ and $C+D<12$. If two single-digit natural numbers add up to $11$ or less, what are the possibilities? (Note that the answer will depend on whether $0$ counts as a natural number.)

The second question is harder. Because it’s multiple choice, however, you can solve it by elimination. First, if you set $x=y=z=\frac13$, you find that $xy+yz+zx=\frac13$; assuming that one of the answers actually does give the set of possible values, this rules out choices (A) and (B), since they both exclude $\frac13$ from the set of possible values. Next, note that

$$1=1^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\;,$$ so

$$xy+yz+zx=\frac12\Big(1-(x^2+y^2+z^2)\Big)\;.$$

Now $x^2+y^2+z^2$ is certainly positive, so $1-(x^2+y^2+z^2)<1$, and therefore

$$xy+yz+zx<\frac12\;.$$

Answer (D) therefore includes too much to be the exact set of possible values: it includes $\frac12$, which we now know isn’t possible. Thus, only (C) can be the exact set of possible values of $xy+yz+zx$.

In fact this can be proved by showing that $x^2+y^2+z^2\ge\frac13$ when $x+y+z=1$ and $x,y$, and $z$ are positive, but I don’t immediately see a way to do that without using a bit more than what I consider precalculus algebra.

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For the inequality note that:

$$(x-y)^2 + (y-z)^2+(z-x)^2 = 2(x^2+y^2+z^2)-2(xy+yz+zx) \geq 0$$

so that $$x^2+y^2+z^2 \geq xy+yz+zx$$

Now

$$1=(x+y+z)^2=(x^2+y^2+z^2) + 2(xy+yz+zx)\geq 3(xy+yz+zx)$$

I think that is pre-calculus algebra, and not nearly as well known a trick as it should be.

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Good point! I missed it! – Tapu Jun 25 '12 at 21:07
Three answers and still no up-votes except mine. – Michael Hardy Jun 25 '12 at 23:24

Brian's answer is fine. If we are allowed to use cauchy-schwarz inequality then $$xy+yz+zx\le \sqrt{x^2+y^2+z^2}. \sqrt{x^2+y^2+z^2}=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=1-2(xy+yz+zx)$$

Hence $xy+yz+zx\le \frac{1}{3}$

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