Bugi wrote: Let $ a,b,c$ be positive real numbers such that $ abc = \frac {9}{4}$. Prove the inequality: $ a^3 + b^3 + c^3 > a\sqrt {b + c} + b\sqrt {c + a} + c\sqrt {a + b}$ Jury's variant: Prove the same, but with $ abc = 2$ (I'll just prove Jury's variant. The other follows directly from it) Obviously $ a^3 + b^3 + c^3 \ge 6$ by AM-GM. So: $ a^3 + b^3 + c^3 \ge \sqrt {6(a^3 + b^3 + c^3)} \ge \sqrt {2(a^2 + b^2 + c^2)(a + b + c)}$ By rearrangement, AM-GM, muirhead, something... And $ \sqrt {(a^2 + b^2 + c^2)((b + c) + (c + a) + (a + b))} \ge a\sqrt {b + c} + b\sqrt {a + c} + c\sqrt {a + b}$ and it is proved.

\[ \a^{3}+b^{3}%Error. "geab" is a bad command. (a+b)\] By AM-GM inequality\[ a^{3}+b^{3}+c^{3}%Error. "geab" is a bad command. (a+b)+c^{3}\ge2\sqrt{abc^{3}(a+b)}=3c\sqrt{a+b}\]. $ =>$ $ =>$ a^{3}+b^{3}+c^{3}\gebc(b+c)+a^{3}\ge2\sqrt{cba^{3}(b+c)}=3a\sqrt{b+c}\[ and a^{3}+b^{3}+c^{3}%Error. "geac" is a bad command. (a+c)+b^{3}\ge2\sqrt{acb^{3}(a+c)}=3b\sqrt{a+c}\] $ =>$ $ =>$ $ 3(a^{3}+b^{3}+c^{3})\ge3(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b})$ DONE!!!

I`m sorry. $ a^{3}+b^{3}%Error. "geab" is a bad command. (a+b)$ $ a^{3}+b^{3}+c^{3}%Error. "geab" is a bad command. (a+b)+c^{3}\ge2\sqrt{abc^{3}(a+b)}=3c\sqrt{a+b}$ $ =>$ $ 3(a^{3}+b^{3}+c^{3})\ge3(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b})$ DONE!

Denote $LHS=a^{3}+b^{3}+c^{3},RHS=a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$ $LHS=\frac{a^{3}+b^{3}}{2}+\frac{b^{3}+c^{3}}{2}+\frac{c^{3}+a^{3}}{2} \geq \frac{ab(a+b)}{2}+\frac{bc(b+c)}{2}+\frac{ac(a+c)}{2}=\frac{9}{8} (\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}) > \frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b} \geq \frac{RHS^{2}}{LHS}$ From here $LHS^{2}>RHS^{2}$ and $LHS>RHS$ Q.E.D

Bugi wrote: Let $ a,b,c$ be positive real numbers such that $ abc=\frac{9}{4}$. Prove the inequality: $ a^3 + b^3 + c^3 > a\sqrt {b + c} + b\sqrt {c + a} + c\sqrt {a + b}$ Jury's variant: Prove the same, but with $ abc=2$ Also we can prove same, but with $abc=1$. use $ab(a+b)\leq a^3+b^3$.

dr_Civot wrote: Bugi wrote: Let $ a,b,c$ be positive real numbers such that $ abc=\frac{9}{4}$. Prove the inequality: $ a^3 + b^3 + c^3 > a\sqrt {b + c} + b\sqrt {c + a} + c\sqrt {a + b}$ Jury's variant: Prove the same, but with $ abc=2$ Also we can prove same, but with $abc=1$. use $ab(a+b)\leq a^3+b^3$. Actually we can't... For $a=b=c=1$ we have to prove that $3>3\sqrt{2}$ which apparently is not

For abc = 2,we can prove it like this.First by AM-GM we have a^3+b^3 >= ab(a+b) (a^3+a^3+b^3 >= 3*a^2*b and b^3+b^3+a^3 >= 3*b^2*a).Then ab(a+b)+2*c^3 >= 2*sqrt(abc*2*c^2*(a+b)) = 4*c*(sqrt(a+b)) and then cyclic .

By Cauchy's inequality, $ a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} \le\ \sqrt{2(a+b+c)(a^2+b^2+c^2)} $ so we have to prove that $ a^3+b^3+c^3 \ge \ \sqrt{2(a+b+c)(a^2+b^2+c^2)} $. By Chebyshev, $ a^3+b^3+c^3\ge\ \frac{(a+b+c)(a^2+b^2+c^2)}{3} $, so $ a^3+b^3+c^3\ge\ \frac{(a+b+c)(a^2+b^2+c^2)}{3} \ge\ \sqrt{2(a+b+c)(a^2+b^2+c^2)} $ or $ (a+b+c)(a^2+b^2+c^2) \ge \ 18 $ which is true since $ (a+b+c)(a^2+b^2+c^2) \ge\ 9abc=\frac{81}{4}>18 $

For $ abc=2 $ it's very easy. By $AM-GM$ we prove that $ \frac{a^3+b^3}{4}\ge \frac{ab(a+b)}{4}=\frac{2ab(a+b)}{4*abc}=\frac{(a+b)}{2c}.$ So again by $AM-GM$ $\frac{a^3+b^3}{4} +\frac{c^3}{2}\ge\frac{(a+b)}{2c}+\frac{c^3}{2}\ge c\sqrt{a+b}$ and by repeating this we get the wanted inequality. $\Box$

Another approach with AM-GM Schur's inequality $a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}=a\sqrt{abc(\frac{b+c}{2})}+b\sqrt{abc(\frac{a+c}{2})}+c\sqrt{abc(\frac{a+b}{2})}\leq(AM-GM)\leq\frac{3abc+ \frac{ab^2+ac^2+bc^2+b^2c+a^2b+a^2c}{2}}{2}\leq(Schur's)\leq \frac{3abc+\frac{a^3+b^3+c^3+3abc}{2}}{2}\leq(AM-GM)\leq\frac{a^3+b^3+c^3+\frac{a^3+b^3+c^3+a^3+b^3+c^3}{2}}{2}=a^3+b^3+c^3$ as desired. Equality occurs when $a=b=c=\sqrt[3]{2}$.