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Thread: 2 Inequality Proofs

  1. #1
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    2 Inequality Proofs

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

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

    2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ 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 $\displaystyle \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geq a+b+c$ given that a, b, and c are positive real numbers.

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

    2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ 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 $\displaystyle (u-v)^2 + (v - w)^2 + (w-u)^2 \geq 0$ gives us: $\displaystyle u^2 + v^2 + w^2 \geq uv + vw + wu. \ \ \ \ (1)$

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

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

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

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

    2.

    I'll use the lemma here. (1)

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

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

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

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

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

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

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

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

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

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

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

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

    2. Prove that $\displaystyle \frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$ 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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