Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer
    
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 $xy+yz+zx\le \frac{1}{3}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.