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ivanwangsa

Bilangan real dengan jumlah nol

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Misalkan $x,y,z$ adalah bilangan real sehingga $x+y+z=0$. Buktikan bahwa


$$\frac{x(x+2)}{2x^2+1} + \frac{y(y+2)}{2y^2+1} + \frac{z(z+2)}{2z^2+1}\geq 0$$


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Perhatikan bahwa :
$\sum \frac{x(x+2)}{2x^2+1}=\sum ({\frac{(2x+1)^2}{2(2x^2+1)}-\frac{1}{2}}).$
Sehingga cukup dibuktikan 
$\sum \frac{(2x+1)^2}{2(2x^2+1)} \ge \frac{3}{2}$ , ekuivalen dengan $\sum \frac{(2x+1)^2}{2x^2+1} \ge 3$
Menurut CS, kita punya
$2x^2+1=\frac{4}{3}x^2+\frac{2}{3}(y+z)^2+1 \le \frac{4}{3}x^2+\frac{4}{3}(y^2+z^2)+1.$
sehingga diperoleh
$\sum \frac{(2x+1)^2}{2x^2+1} \ge 3\sum \frac{(2x+1)^2}{4(x^2+y^2+z^2)+3}=3.$
terbukti.

Edited by theoneandonly
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