work 024 – mathlinks user wya

Problem

Let a,b,c be positive real numbers. Prove that

\frac {a^{2} + bc}{b^{2} + ca} + \frac {b^{2} + ca}{c^{2} + ab} + \frac {c^{2} + ab}{a^{2} + bc}\le\frac {b}{a} + \frac {c}{b} + \frac {a}{c}.

Proof

\sum \frac {a^2 + bc}{b^2 + ca} = \sum \frac {\frac {a^2}{c^2} + \frac {b}{c} }{\frac {b^2}{c^2} + \frac {a}{c}}
Let \frac {b}{a} = x , \cdots
therefore , xyz = 1
we have to prove that –
x + y + z \ge \sum x \frac {z + y}{y + x}

\Leftrightarrow \sum x \cdot \frac {(x - z)}{x + y} \ge 0

\Leftrightarrow \sum (xy+xz)(x^2-z^2) \ge 0

\Leftrightarrow x^3y + y^3z + z^3y \ge xyz(x + y + z)

\Leftrightarrow \sum \frac {x^2}{y} \ge x + y + z
which is true by C-S 🙂

Equality hold for $x=y=z$ .

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One comment

  1. Incidentally … this also gives us one more property of the G function by Pham Kim Hung 🙂 –
    G(b,a,c) \ge G(a^2+bc,b^2+ca,c^2+ab)

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