enneper wrote: 4. Let $a \geq b \geq c \geq d>0$. Show that \[ \frac{b^3}{a} + \frac{c^3}{b} + \frac{d^3}{c} + \frac{a^3}{d} + 3 \left( ab+bc+cd+da \right) \geq 4 {\left( a^2 + b^2 + c^2 +d^2 \right)}. \]Other problems (in Korean) are also available at https://www.facebook.com/KoreanMathOlympiad Let $a \geq b \geq c >0$. Show that \[ \frac{b^3}a+\frac{c^3}b+\frac{a^3}c+ ab+bc+ca\ge 2 ( a^2 + b^2 + c^2 ) \]

sqing wrote: enneper wrote: 4. Let $a \geq b \geq c \geq d>0$. Show that \[ \frac{b^3}{a} + \frac{c^3}{b} + \frac{d^3}{c} + \frac{a^3}{d} + 3 \left( ab+bc+cd+da \right) \geq 4 {\left( a^2 + b^2 + c^2 +d^2 \right)}. \]Other problems (in Korean) are also available at https://www.facebook.com/KoreanMathOlympiad Let $a \geq b \geq c >0$. Show that \[ \frac{b^3}a+\frac{c^3}b+\frac{a^3}c+ ab+bc+ca\ge 2 ( a^2 + b^2 + c^2 ) \] By AM-GM inequality $LHS=\sum_{cyc} (\frac{b^3}{a}+ab)\geq \sum_{cyc} 2b^2=2(a^2+b^2+c^2) $

NahTan123xyz wrote: sqing wrote: enneper wrote: 4. Let $a \geq b \geq c \geq d>0$. Show that \[ \frac{b^3}{a} + \frac{c^3}{b} + \frac{d^3}{c} + \frac{a^3}{d} + 3 \left( ab+bc+cd+da \right) \geq 4 {\left( a^2 + b^2 + c^2 +d^2 \right)}. \]Other problems (in Korean) are also available at https://www.facebook.com/KoreanMathOlympiad Let $a \geq b \geq c >0$. Show that \[ \frac{b^3}a+\frac{c^3}b+\frac{a^3}c+ ab+bc+ca\ge 2 ( a^2 + b^2 + c^2 ) \] By AM-GM inequality $LHS=\sum_{cyc} (\frac{b^3}{a}+ab)\geq \sum_{cyc} 2b^2=2(a^2+b^2+c^2) $ Thanks. Let $a, b , c >0$. Then \[ \frac{b^3}a+\frac{c^3}b+\frac{a^3}c+ ab+bc+ca\ge 2 ( a^2 + b^2 + c^2 ) \] Let $a \geq b \geq c >0$. Show that \[ \frac{b^3}a+\frac{c^3}b+\frac{a^3}c+ 2(ab+bc+ca)\ge 3 ( a^2 + b^2 + c^2 ) \]

enneper wrote: 4. Let $a \geq b \geq c \geq d>0$. Show that \[ \frac{b^3}{a} + \frac{c^3}{b} + \frac{d^3}{c} + \frac{a^3}{d} + 3 \left( ab+bc+cd+da \right) \geq 4 {\left( a^2 + b^2 + c^2 +d^2 \right)}. \]Other problems (in Korean) are also available at https://www.facebook.com/KoreanMathOlympiad $LHS-RHS=\sum_{cyc} (\frac{b^3}{a}-3b^2+3ba-a^2)=\sum_{cyc} \frac{(b-a)^3}{a}$ but $(a-d)^3\geq (a-b)^3+(b-c)^3+(c-a)^3$ So $\sum_{cyc} \frac{(b-a)^3}{a}\geq (a-b)^3(\frac{a-d}{ad})+(b-c)^3(\frac{b-d}{bd})+(c-d)^3(\frac{c-d}{cd})\geq 0$ Done : )

NahTan123xyz wrote: enneper wrote: 4. Let $a \geq b \geq c \geq d>0$. Show that \[ \frac{b^3}{a} + \frac{c^3}{b} + \frac{d^3}{c} + \frac{a^3}{d} + 3 \left( ab+bc+cd+da \right) \geq 4 {\left( a^2 + b^2 + c^2 +d^2 \right)}. \]Other problems (in Korean) are also available at https://www.facebook.com/KoreanMathOlympiad $LHS-RHS=\sum_{cyc} (\frac{b^3}{a}-3b^2+3ba-a^2)=\sum_{cyc} \frac{(b-a)^3}{a}$ but $(a-d)^3\geq (a-b)^3+(b-c)^3+(c-a)^3$ So $\sum_{cyc} \frac{(b-a)^3}{a}\geq (a-b)^3(\frac{a-d}{ad})+(b-c)^3(\frac{b-d}{bd})+(c-d)^3(\frac{c-d}{cd})\geq 0$ Done : ) Very nice.

Nice one We want to prove $A = \sum \frac{b^3}{a}+3(\sum ab) - 4(\sum a^2) \geq 0$ . Also we can see $\frac{b^3}{a}+3ab-3b^2-a^2 = \frac{(b-a)^3}{a}$ . Then we can see $A = \sum \frac{(b-a)^3}{a}$ . Now we need to prove $\sum \frac{(b-a)^3}{a} \geq 0$ . Define $a-b = x , b-c = y , c-d = z $ . We can see $ a-d=x+y+z$ . We need to prove $\frac{(x+y+z)^3}{d} \geq \sum \frac{x^3}{a}$ . By $a \geq b \geq c \geq d>0$ we can see $$\frac{(x+y+z)^3}{d} \geq \sum_{x,y,z} \frac{x^3}{d}$$ Now we need to prove $\sum _{x,y,z}\frac{x^3}{d} \geq \sum_{a,b,c} \frac{x^3}{a}$ . Now ....... ? $\blacksquare$ .