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ข้อความเดิมเขียนโดยคุณ LightLucifer

ALGEBRA NO.7
ให้ $x,y,z>0$ โดยที่ $x+y+z=1$ จงพิสูจน์ว่า
$$\sum_{cyc}^{}\frac{x^4}{(x+2y)(y+2x)}\geqslant \frac{1}{27}$$

Typical Approach by Cauchy-Schwarz Inequality:

LHS $\geq \dfrac{(x^2+y^2+z^2)^2}{4(x^2+y^2+z^2)+5(xy+yz+zx)}\geq\dfrac{1}{27}$

$\because 3(x^2+y^2+z^2)\geq (x+y+z)^2\Rightarrow 3(x^2+y^2+z^2)\geq 1$

$\therefore 12(x^2+y^2+z^2)^2\geq 4(x^2+y^2+z^2)$

$15(x^2+y^2+z^2)^2\geq 5(x^2+y^2+z^2)\geq 5(xy+yz+zx)$

$27(x^2+y^2+z^2)^2 \geq 4(x^2+y^2+z^2)+5(xy+yz+zx)$