Results 1 to 4 of 4

Math Help - Proving in algebra?

  1. #1
    Junior Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    74

    Question Proving in algebra?

    can someone teach me how to prove algebraic problems? like this one,
    show that if the sum of two positive numbers is 1, then the sum of their squares is ≥ 1/2 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Kaloda View Post
    can someone teach me how to prove algebraic problems? like this one,
    show that if the sum of two positive numbers is 1, then the sum of their squares is ≥ 1/2 ?

    x+y=1\Longrightarrow y=1-x\,,\,\,x^2+y^2=(x+y)^2-2xy=1^2-2xy=1-2xy\geq 1-\frac{1}{2}=\frac{1}{2}

    Now, the last equality is true if  2xy\leq \frac{1}{2}\Longleftrightarrow xy\leq \frac{1}{4} , so now just substitute y=1-x here , get a

    quadratic inequality in x and check that iit is always true...

    Hmmm...if you're a beginner the above may be a little hard (if you're in high school you shoud be able to do it, though), so perhaps there's

    an easier way to do it WITHOUT the use of derivatives.

    Also remember than in english, just as in spanish, german, russian, italian, etc., when you begin a sentence after a

    period you must use capital letters.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Kaloda View Post
    can someone teach me how to prove algebraic problems? like this one,
    show that if the sum of two positive numbers is 1, then the sum of their squares is ≥ 1/2 ?
    x+y=1 gives the relationship between x and y.

    x^2+y^2 is the sum of the squares and we want it's minimum value.

    Use x+y=1 to write the sum of squares in terms of x only.

    x+y=1\Rightarrow\ y=1-x,\;\;\;x<1,\;\;y<1

    x^2+y^2=x^2+(1-x)^2=x^2+1-2x+x^2=2x^2-2x+1

    This is a U-shaped quadratic in x, valid for 0<x<1
    The parabola is symmetrical about the minimum.

    The minimum lies midway between the roots of the quadratic, even if the roots are complex,
    which in this case they are as the graph lies entirely above the x-axis.

    The roots of ax^2+bx+c=0 are

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    The minimum lies midway between these, hence it lies at x=-\frac{b}{2a}=-\frac{-2}{2(2)}=\frac{1}{2}

    \left(\frac{1}{2}\right)^2+\left(1-\frac{1}{2}\right)^2=\frac{1}{4}+\frac{1}{4}

    Therefore x^2+y^2\ \ge\ \frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    A shortcut would be to find two equal values for 2x^2-2x+1

    then find midway between them.
    We can find f(x)=1

    2x^2-2x+1=1\Rightarrow\ 2x^2-2x=0\Rightarrow\ 2x(x-1)=0
    \Rightarrow\ x=1,\;\;\;x=0

    Midway between these is x=\frac{1}{2}

    Therefore f\left(\frac{1}{2}\right) gives the minimum.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 6th 2010, 04:03 PM
  2. Replies: 0
    Last Post: April 24th 2010, 12:37 AM
  3. [SOLVED] Proving k*0=0 using linear algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 15th 2010, 03:11 PM
  4. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 02:30 PM
  5. Proving unions by Algebra of Sets
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: November 18th 2007, 10:35 AM

Search Tags


/mathhelpforum @mathhelpforum