any solution?

The solution was given by my friend and I during the camp as we were members in the camp. Let $x=\dfrac{2a}{b+c}, y=\dfrac{2b}{a+c}, z=\dfrac{2c}{a+b}$. After the substitution, we will have $LHS=\sum \sqrt{\dfrac{xy+x+y}{z}}=\sum \sqrt{\dfrac{xyz+xz+yz}{z^2}}=\sum\sqrt{\dfrac{4-xy}{z^2}}=\sum(a+b)\sqrt{\dfrac{a+b+c}{c(a+c)(b+c)}}\geq 3\sqrt[3]{\sqrt{\dfrac{(a+b+c)^3}{abc}}}=3\sqrt{a+b+c}\dfrac{1}{\sqrt[6]{abc}}$ $RHS=3\sqrt{3}\sqrt{\prod\dfrac{2(a+b+c)}{4a+b+c}}$ We want to prove $LHS\geq RHS$, since it is homogenous, we can let $a+b+c=3$ After simplification, our target is: If $a+b+c=3$, prove$(1+a)(1+b)(1+c)\geq 8\sqrt[3]{abc}$ Though the inequality looks beautiful, we still didn't have a nice proof to the inequality. The way I used was Lagrange multiplier. Just consider:$f(a,b,c)=\sum {ln(1+a)-\dfrac{1}{3}ln(a)}$ , $g(a,b,c)=a+b+c$. After solving $\dfrac{\partial(f-\lambda g)}{\partial a}=\dfrac{\partial (f-\lambda g)}{\partial y}=\dfrac{\partial (f-\lambda g)}{\partial z}=0$,we'll get $a=b=c $and$ \lambda=ln 2$.

supercarry wrote: After simplification, our target is: If $a,b,c>0$ and $a+b+c=3$, prove$(1+a)(1+b)(1+c)\geq 8\sqrt[3]{abc}$ This inequality is surprisingly hard. I could not think of any clean ways to finish it ---most of them involve setting two variables to be the same (say, via uvw, HCF or pigeonhole). I'm interested in seeing a good way to prove this.

Instead of using Lagrange multiplier, for the final step you could check that $\log(1+a) - \frac 1 3 \log a$ is convex below $a=\frac{1+\sqrt 3}{2}$ and concave above, and then you have a few cases to check for equality (either one of $a,b,c$ is 0 or at most one is less than $\frac{1+\sqrt 3}{2}$ and the rest are equal). Think this is basically HCF mentioned by USJL.

5/16 is less than 1/3?

DVDthe1st wrote:

5/16 is less than 1/3?

Nice one!

<DVDthe1st> wrote: Note that $(ab+bc+ca)^2 \ge 3abc(a+b+c)$, so $$(1+a)(1+b)(1+c) \ge 1 + 3 + 3\sqrt{abc} + abc \ge 8(abc)^{5/16} \ge 8(abc)^{1/3}$$since $abc \le 1$. I think it's so outstanding . Actually I have been working on it and I was stuck to make the power into $\dfrac{1}{3}$, quite interesting and perfect. Thanks for giving me a nice way to solve the last inequality!