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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{4ab+5c^2}{a+b}+\frac{4bc+5a^2}{b+c}+\frac{4ca+5b^2}{c+a}\ge \frac{3}{2}\cdot\frac{(a+b+c)^3}{ab+bc+ca}.
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