อสมการครับ

[FranceZii Siriseth]

$$\sum_{cyc}\frac{a^2}{b+c+d} \geqslant \frac{4}{3} , a^2+b^2+c^2+d^2=4 $$

[จูกัดเหลียง]

Consider# from the power-mean $$ \Big(\dfrac{\sum a^3}{4}\Big)^{1/3}\ge \Big(\frac{\sum a^2}{4}\Big)^{1/2}=1$$
or equivalent to $\displaystyle \sum_{cyc} a^3\ge 4$ and in the same we get $\displaystyle \sum_{cyc} a\le 4...*$
and consider the Cauchy $$\sum_{cyc} \frac{a^2}{b+c+d}=\sum_{cyc}\frac{a^4}{a^2(b+c+d)}\ge\frac{(a^2+b^2+c^2+d^2)^2}{\sum a(b^2+c^2+d^2)}=\frac{16}{4\sum a-\sum a^3 }$$
from $*$ So $RHS.\ge \dfrac{16}{4\cdot4-4}=\dfrac{4}{3}$
as desired

[FranceZii Siriseth]

ข้อความเดิมเขียนโดยคุณ จูกัดเหลียง

Consider# from the power-mean $$ \Big(\dfrac{\sum a^3}{4}\Big)^{1/3}\ge \Big(\frac{\sum a^2}{4}\Big)^{1/2}=1$$
or equivalent to $\displaystyle \sum_{cyc} a^3\ge 4$ and in the same we get $\displaystyle \sum_{cyc} a\le 4...*$
and consider the Cauchy $$\sum_{cyc} \frac{a^2}{b+c+d}=\sum_{cyc}\frac{a^4}{a^2(b+c+d)}\ge\frac{(a^2+b^2+c^2+d^2)^2}{\sum a(b^2+c^2+d^2)}=\frac{16}{4\sum a-\sum a^3 }$$
from $*$ So $RHS.\ge \dfrac{16}{4\cdot4-4}=\dfrac{4}{3}$
as desired

ขอบคุณมากครับ ลืมใช้ power mean