Steve12345 wrote: For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove: $\sum \frac{a}{\sqrt{a^3+5}} \leq \frac{\sqrt{6}}{2}$ Observe that $2(a^3+5)=(a^3+a^3+1)+9\geq 3a^2+9=3(a+b)(a+c)$. Hence, we have $$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}}$$ $$=\sqrt{\frac{2}{3}\cdot \frac{\displaystyle{\left(\sum_{cyc} \sqrt{a^2(b+c)}\right)^2}}{(a+b)(b+c)(c+a)}}$$ (By CS.) $$\leq \sqrt{\frac{2}{3}\cdot \frac{(\sum a)(\sum (ab+ac))}{(a+b)(b+c)(c+a))}}\leq \frac{\sqrt{6}}{2}$$Done

Steve12345 wrote: For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove: $\sum \frac{a}{\sqrt{a^3+5}} \leq \frac{\sqrt{6}}{2}$ This problem was proposed by me, it was A6.

I observe that $f''(x)=\frac{x}{\sqrt{x^3+5}}<0$, so we can use Jensen as well.

minageus wrote: I observe that $f''(x)=\frac{x}{\sqrt{x^3+5}}<0$, so we can use Jensen as well. It is not true.

In the interval $x\in (0,3]$ it is true. Try using sumbolab.

Where can we see the whole shortlist?

I'll post some more the day after tomorrow.

NahTan123xyz wrote: Steve12345 wrote: For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove: $\sum \frac{a}{\sqrt{a^3+5}} \leq \frac{\sqrt{6}}{2}$ Observe that $2(a^3+5)=(a^3+a^3+1)+9\geq 3a^2+9=3(a+b)(a+c)$. Hence, we have $$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}}$$ $$=\sqrt{\frac{2}{3}\cdot \frac{\displaystyle{\left(\sum_{cyc} \sqrt{a^2(b+c)}\right)^2}}{(a+b)(b+c)(c+a)}}$$ (By CS.) $$\leq \sqrt{\frac{2}{3}\cdot \frac{(\sum a)(\sum (ab+ac))}{(a+b)(b+c)(c+a))}}\leq \frac{\sqrt{6}}{2}$$Done Nice solution! Another way to finish would be $$\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}} \leq \frac{1}{2} \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{3}{2}$$from AM-GM.

Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3 .$ Then $$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}} \leq \frac{1}{2} \sqrt{\frac{2}{3}}\sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{\sqrt{6}}{2}$$

Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3.$ Prove that $$\frac{a}{\sqrt{a^{n}+2n-1}} +\frac{b}{\sqrt{b^{n}+2n-1}}+\frac{c}{\sqrt{c^{n}+2n-1}}\leq \frac{3}{\sqrt{2n}}.$$Where $2\leq n\in N^+.$

@above Solution Please

XbenX wrote: NahTan123xyz wrote: Steve12345 wrote: For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove: $\sum \frac{a}{\sqrt{a^3+5}} \leq \frac{\sqrt{6}}{2}$ Observe that $2(a^3+5)=(a^3+a^3+1)+9\geq 3a^2+9=3(a+b)(a+c)$. Hence, we have $$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}}$$ $$=\sqrt{\frac{2}{3}\cdot \frac{\displaystyle{\left(\sum_{cyc} \sqrt{a^2(b+c)}\right)^2}}{(a+b)(b+c)(c+a)}}$$ (By CS.) $$\leq \sqrt{\frac{2}{3}\cdot \frac{(\sum a)(\sum (ab+ac))}{(a+b)(b+c)(c+a))}}\leq \frac{\sqrt{6}}{2}$$Done Nice solution! Another way to finish would be $$\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}} \leq \frac{1}{2} \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{3}{2}$$from AM-GM. Why does the following hold? $\frac{1}{2} \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{3}{2}$

Nevermind that was a stupid question ( I was confused about the inequality, it is just an equality...)

bora_olmez wrote: Why does the following hold? $\frac{1}{2} \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{3}{2}$ $$ \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) = \sum_{cyc} \left( \frac{a}{a+b}+\frac{b}{b+a} \right) =3$$

sqing wrote: bora_olmez wrote: Why does the following hold? $\frac{1}{2} \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) \leq \frac{3}{2}$ $$ \sum_{cyc} \left( \frac{a}{a+b}+\frac{a}{a+c} \right) = \sum_{cyc} \left( \frac{a}{a+b}+\frac{b}{b+a} \right) =3$$ Yes, I realized after asking the question (I was confused because it said less than or equal to 3) Thanks, anyways...