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Inequality Problem

  1. Apr 29, 2012 #1
    5. Suppose that x, y and z are positive real numbers such that xyz = 1.
    (a) Prove that
    27 [itex]\leq[/itex](1 + x + y)[itex]^{2}[/itex] + (1 + y + z)[itex]^{2}[/itex] + (1 + z + x)[itex]^{2}[/itex]

    with equality if and only if x = y = z = 1.
    (b) Prove that
    (1 + x + y)[itex]^{2}[/itex] + (1 + y + z)[itex]^{2}[/itex] + (1 + z + x)[itex]^{2}[/itex] [itex]\leq[/itex] 3(x + y + z)[itex]^{2}[/itex]

    with equality if and only if x = y = z = 1.
    3. The attempt at a solution
    I don't really how to prove this. I can visualize the truth in my head but I don't know where to start a proof. Does anyone know any particular methods to solve symmetric equalities such as this one? I'm trying to see a nice way to simplify it or introduce a substitution but i can't see anything. Thanks
     
  2. jcsd
  3. Apr 30, 2012 #2
    You know that xyz=1 so that would be a good start. And also the sqrt could be taken on each side and then you just subsittute till you have something which you know is greater than [itex]\sqrt{27}[/itex]
     
  4. Apr 30, 2012 #3
    Hints:

    5. (a) Rewriting the expressions a bit, can you equivalently prove that [tex](x+1)^2 + (y+1)^2 + (z+1)^2 + xy + yz + zx \geq 15[/tex]

    5. (b) Put [tex] x + y + z = S_1[/tex]
    [tex] xy + yz + zx = S_2[/tex]

    and rewrite the expressions.

    Now can you equivalently prove that
    [tex] 2S_2 \geq 4S_1 + 3 - S_1^2[/tex]
     
  5. Apr 30, 2012 #4
    ok i can see how you rearranged the first inequality but i still don't know where to go from there ? ? ? And may i ask where the inspiration came from to arrange the inequality in that way? Sorry, i dont have much experience solving this kind of problem, so any tips would help immensely! Thanks
     
  6. Apr 30, 2012 #5
    The idea is to apply the AM-GM-HM inequality at some stage or the other. If you keep that in mind, you can try to manipulate the expressions accordingly.

    Can you see that


    [tex](x+1)^2 \geq 4x[/tex]
    [tex](y+1)^2 \geq 4y[/tex]
    [tex](z+1)^2 \geq 4z[/tex]

    Can you try to complete the proof now? Let me know if you need more help.
     
  7. Apr 30, 2012 #6
    Oh yes i can see, it comes from the fact that (x-1)^2 ≥ 0
    So logically it seems to me to try prove

    4x + 4y + 4z + xy + yz +xz ≥ 15

    Applying AM-GM we get :

    (4x + 4y + 4z) ≥ 12

    So now we have

    xy + yz + zx ≥ 3

    which is equivalent to

    1/x + 1/y + 1/z ≥ 3

    Now I am stuck :P
     
  8. Apr 30, 2012 #7
    A simpler way to complete the proof is to use the following facts:
    [tex] x + y + z \geq 3 [/tex] and [tex] xy + yz + zx \geq 3 [/tex]

    These facts follow from the AM-GM-HM inequality:

    [tex] \frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

    Which means, [tex] \frac{x+y+z}{3} \geq 1 \geq \frac{3}{xy + yz + zx}[/tex]

    [From the above line, your required proof also follows immediately!]
     
  9. Apr 30, 2012 #8
    I tried doing that, however i got 2S_1 rather than 4S_1, did you make an error calculating or was that me?
     
  10. Apr 30, 2012 #9
    Please check my calculations. I did it in a hurry.

    We are required TPT [tex] (S_1 - x + 1)^2 + (S_1 - y + 1)^2 + (S_1 - z + 1)^2 \leq 3S_1^2[/tex]

    Simplifying, we find that we are required TPT
    [tex] S_1^2 - 2S_2 \leq 2S_1^2 - 4S_1 - 3 [/tex]
    i.e., TPT [tex] 2S_2 \geq 4S_1 + 3 - S_1^2[/tex]
     
  11. May 1, 2012 #10
    sorry, yes, you're completely right, i must have made a mistake. Ok so is this logic ok?
    2S2 ≥ 3 + S1(4-S1)
    Well we already know that S1≥3 and S2 ≥ 3
    So 2S2 ≥ 6
    On the other hand, if we wish to maximise the right hand side of the equation, we want S1 to be its minimum, 3, hence the maximum value of the r.h.s. = 3 + 3(4-3) =6
    So minimum l.h.s. = 6, max r.h.s. = 6, hence l.h.s. ≥ r.h.s
    Equality occurs if x = y = z = 1, because at these values, the min and max, respectively, occur.
    Is that rigorous enough a proof? I still didnt manage to prove that equality occurs if and only if x =y =z =1? Thanks
     
  12. May 1, 2012 #11
    Observe that
    [tex] 4S_1 + 3 - S_1^2 = 7 - (S_1 - 2)^2 [/tex]

    Since [tex] S_1 \geq 3[/tex] the RHS is always [tex] \leq 6 [/tex]

    Finally the LHS is always [tex] \geq 6 [/tex] and the RHS is always [tex] \leq 6 [/tex], which completes the proof.

    You can solve the equality case by considering the equations [tex] LHS = RHS = 6 [/tex]
     
  13. May 1, 2012 #12
    Ah ok cool, I had tried factoring and leaving a seven outside too before going back and corssing out my work, but I think be both came to the same conclusion. Many thanks for your help dibyendu!
     
  14. May 1, 2012 #13
    You are most welcome, Evansmiley. It's my pleasure.
     
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