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Math Help - 2 Inequality Proofs

  1. #1
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    2 Inequality Proofs

    1. Prove that \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c given that a, b, and c are positive real numbers.

    ----------------------------

    2. Prove that \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z given that x, y, and z are positive real numbers.
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  2. #2
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    Quote Originally Posted by Winding Function View Post
    1. Prove that \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c given that a, b, and c are positive real numbers.

    ----------------------------

    2. Prove that \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z given that x, y, and z are positive real numbers.
    the questions you've been asking do not belong to "Advanced Algebra" forum. we have a forum for inequalities. anyway, here's a solution to your problems:


    expanding (u-v)^2 + (v - w)^2 + (w-u)^2 \geq 0 gives us: u^2 + v^2 + w^2 \geq uv + vw + wu. \ \ \ \ (1)

    therefore u^4 + v^4 + w^4 \geq (uv)^2 + (vw)^2 + (wu)^2. but again by (1): (uv)^2 + (vw)^2 + (wu)^2 \geq uvw(u+v+w). thus: u^4 + v^4 + w^4 \geq uvw(u+v+w). \ \ \ \ (2)

    now in (1) put u=ab, \ v =bc, \ w=ac and then divide by abc to prove your first problem and in (2) put u=x, \ v = y, \ w=z and then divide by xyz to prove your second problem. \Box
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  3. #3
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    1.

    By Cauchy-Schwarz: <br />
\left( {\tfrac{{bc}}<br />
{a} + \tfrac{{ac}}<br />
{b} + \tfrac{{ab}}<br />
{c}} \right) \cdot \left( {\tfrac{{ab}}<br />
{c} + \tfrac{{bc}}<br />
{a} + \tfrac{{ac}}<br />
{b}} \right) \geqslant \left( {a + b + c} \right)^2 <br />
- since <br />
\left( {\tfrac{{bc}}<br />
{a} \cdot \tfrac{{ab}}<br />
{c} = b^2 ;...} \right)<br />
-

    Thus: <br />
\left( {\tfrac{{bc}}<br />
{a} + \tfrac{{ac}}<br />
{b} + \tfrac{{ab}}<br />
{c}} \right)^2  \geqslant \left( {a + b + c} \right)^2 <br />
and by the positivity of both sums we have: <br />
\tfrac{{bc}}<br />
{a} + \tfrac{{ac}}<br />
{b} + \tfrac{{ab}}<br />
{c} \geqslant a + b + c<br />

    2.

    I'll use the lemma here. (1)

    Using it we have: <br />
\left( {\tfrac{{x^3 }}<br />
{{yz}} + \tfrac{{y^3 }}<br />
{{xz}} + \tfrac{{z^3 }}<br />
{{xy}}} \right) \cdot \left( {yz + xz + xy} \right) \cdot \left( {1 + 1 + 1} \right) \geqslant \left( {x + y + z} \right)^3 <br />

    Now since <br />
x^2  + y^2  + z^2  \geqslant yz + xz + xy<br />
see in the link in (1), we have <br />
x^2  + y^2  + z^2  + 2 \cdot \left( {yz + xz + xy} \right) \geqslant 3 \cdot \left( {yz + xz + xy} \right)<br />
i.e. <br />
\left( {x + y + z} \right)^2  \geqslant 3 \cdot \left( {yz + xz + xy} \right)<br />

    Thus we have: <br />
\left( {\tfrac{{x^3 }}<br />
{{yz}} + \tfrac{{y^3 }}<br />
{{xz}} + \tfrac{{z^3 }}<br />
{{xy}}} \right) \cdot 3 \geqslant \tfrac{{\left( {x + y + z} \right)^2 }}<br />
{{yz + xz + xy}} \cdot \left( {x + y + z} \right) \geqslant 3 \cdot \left( {x + y + z} \right)<br />
<br />
 \Rightarrow \tfrac{{x^3 }}<br />
{{yz}} + \tfrac{{y^3 }}<br />
{{xz}} + \tfrac{{z^3 }}<br />
{{xy}} \geqslant x + y + z \square<br />
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  4. #4
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    On the first inequality, we can start by using
    x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2=2(x^2y^2+y^2z^2+x^2z^2-x^2yz-xy^2z-xyz^2
    (try expanding the left hand side to see that this is true).
    Thus
    \frac{x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2}{2xyz}=\frac{x^2y^2+y^2z^2+x^2z^2-x^2yz-xy^2z-xyz^2}{xyz}
    =\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}-x-y-z
    But if x,y, and z are all positive, then xyz is positive. And as x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2 is the sum of three squares, it cannot be negative. Thus \frac{x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2}{2xyz}\ge0. From our above equality, then, we have
    \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}-x-y-z\ge0
    \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge{x+y+z}

    --Kevin C.
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  5. #5
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    Quote Originally Posted by Winding Function View Post
    1. Prove that \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c given that a, b, and c are positive real numbers.
    That should be simple enough. Let's see....

    \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c

    a^2b^2+a^2c^2+b^2c^2\geq a^2bc+ab^2c+abc^2

    By symmetry, let c\geq b\geq a. Then:

    a^2c^2+b^2c^2-2abc^2\geq a^2bc+ab^2c-a^2b^2-abc^2

    c^2(a^2+b^2-2ab)\geq ab(ac+bc-ab-c^2)

    c^2(b-a)^2\geq ab(c-b)(a-c)

    We can see that c^2(b-a)^2\geq 0, whereas ab(c-b)(a-c)\leq0 (because a-c\leq 0).

    ----------------------------

    2. Prove that \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z given that x, y, and z are positive real numbers.
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    the questions you've been asking do not belong to "Advanced Algebra" forum. we have a forum for inequalities.
    Where is the forum for inequalities? I don't think these problems fit under the category of "Solving and Graphing Inequalities" for Elementary and Middle School Math Help (http://www.mathhelpforum.com/math-help/inequalities/).

    Also, I don't understand your solution. Could you please indicate which parts correspond to which question? Thanks!
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