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Math Help - Prove this ...

  1. #1
    Super Member dhiab's Avatar
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    Prove this ...

    a,b,x,y are a positif reals and x<y
    Prove :
     <br />
\frac{x}{y}\prec(\frac{ax+by}{bx+ay}\prec\frac{y}{  x}<br />
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  2. #2
    Super Member

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    Hello, dhiab!

    a,b,x,y are positive reals and x<y.

    Prove: .  \frac{x}{y} \:<\:\frac{ax+by}{bx+ay} \;<\:\frac{y}{x}

    \begin{array}{ccccc}\text{We have:} & x \:<\:y \\<br />
\text{Square:} & x^2 \:<\:y^2 \\<br />
\text{Multiply by }b\!: & bx^2 \:<\:by^2 \\<br />
\text{Add }axy\!: & bx^2 + axy \:<\:axy + by^2 \\<br />
\text{Factor:} & x(bx+ay) \:<\:y(ac+by) \\<br />
\text{Hence:} & \qquad\;\; \dfrac{x}{y} \:<\:\dfrac{ax+by}{bx+ay}<br />
\end{array}

    \begin{array}{ccccc}\text{We have:} & x \:<\:y \\<br />
\text{Square:} & x^2 \:<\:y^2 \\<br />
\text{Multiply by }a\!:& ax^2 \:<\:ay^2 \\<br />
\text{Add }bxy\!: & ax^2 + bxy \:<\:bxy + ay^2 \\<br />
\text{Factor:} & x(ax+by) \:<\:y(bx+ay) \\<br />
\text{Hence:} & \dfrac{ax+by}{bx+ay} \:<\:\dfrac{y}{x} \qquad\;\;<br />
\end{array}

    . . . \text{Therefore: }\;\frac{x}{y} \;<\;\frac{ax+by}{bx + ay} \;<\:\frac{y}{x}

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