อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Keehlzver
ข้อ 3. กำหนดให้ $x,y,z > 0$ และ $xyz=1$ จงพิสูจน์ว่า $$\frac{x^5+x^2}{(y+z)^2}+\frac{y^5+y^2}{(z+x)^2}+\frac{z^5+z^2}{(x+y)^2}\geq \frac{3(x^3y+y^3z+z^3x)}{2(x+y+z)}$$
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WLOG $x\ge y\ge z>0$
By Cheby's $$\frac{x^5+x^2}{(y+z)^2}+\frac{y^5+y^2}{(z+x)^2}+\frac{z^5+z^2}{(x+y)^2}=\sum_{cyc} \Big(\frac{x}{y+z}\Big)^2(x^3+1)$$ $$\ge \frac{1}{3}\Big(\Big(\frac{x}{y+z}\Big)^2+\Big(\frac{y}{z+x}\Big)^2+\Big(\frac{z}{x+y}\Big)^2\Big)(x^3+y^3+z^3+3)$$
$$\ge \frac{3}{4}(x^2+y^2+z^2+x+y+z)$$
$\therefore $ จึงต้องการ $$\frac{3}{4}(x^2+y^2+z^2+x+y+z)\ge \frac{3}{2}\Big(\frac{x^3y+y^3z+z^3x}{x+y+z}\Big)$$
$$\leftrightarrow (x+y+z)(x^2+y^2+z^2+x+y+z)\ge 2(x^3y+y^3z+z^3x)$$
but $2(x^3y+y^3z+z^3x)\le \frac{1}{3}(x+y+z)^2(x^2+y^2+z^2)$
It's Remain to show that $(x+y+z)(x^2+y^2+z^2+x+y+z)\ge \frac{1}{3}(x+y+z)^2(x^2+y^2+z^2)...(*)$
Consider $a^2-a+(-3-2b)\le 0 \because D\ge 0$
Deplace $a=x+y+z,b=xy+yz+zx$ Then $(*)$ is true
เเต่ผมไม่ได้ใช้ $xyz=1$ เลยอ่ะครับ
