IstekOlympiadTeam wrote: $a,b,c\in\mathbb{R^+}$ and $a^2+b^2+c^2=48$. Prove that \[a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16}\le24^2\] now this looks cool! [Moderator says: do not quote the entire post before yours]

What about this http://artofproblemsolving.com/community/c6h1084477p4786093

By $AM-GM$ we have : $a^2\sqrt{2b^3+16}=a^2\sqrt{2(b^3+8)}=a^2\sqrt{2(b+2)(b^2-2b+4)} \leq a^2\left(\frac{2(b+2)+(b^2-2b+4)}{2}\right)=\frac{a^2b^2+8a^2}{2}$ Thus $$a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16} \leq \frac{a^2b^2+b^2c^2+c^2a^2+8(a^2+b^2+c^2)}{2}=\frac{a^2b^2+b^2c^2+c^2a^2}{2}+192$$ and $$3(a^2b^2+b^2c^2+c^2a^2) \leq (a^2+b^2+c^2)^2=2304$$ thus $$\frac{a^2b^2+b^2c^2+c^2a^2}{2} \leq 384$$ We get the desired result by adding the last two results

Let $a,b,c\geq 0$ and $a^2+b^2+c^2=48$. Prove that \[a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16}\geq 192\]

sqing wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2=48$. Prove that \[a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16}\geq 192\] This is nothing, just the fact that : $a^2\sqrt{2b^3+16} \geq a^2\sqrt{16}=4a^2$ and the equality doesn't occur ... [Moderator says: do not quote the entire post before yours]

Let $a,b,c\geq 0$ and $a^2+b^2+c^2=3$ . For $\lambda\geq 0$ , prove that \[a^2\sqrt{b^3+\lambda}+b^2\sqrt{c^3+\lambda}+c^2\sqrt{a^3+\lambda}\leq 3\sqrt{1+\lambda}.\]

http://www.artofproblemsolving.com/community/c6h1136041p5307991

Analyzing $f(t) = \sqrt{2t^{\frac{3}{2}}+16}$ at $t=16$, we desire inequality \[ \sqrt{2b^3+16} \le \frac{b^2}{2} +4 \]which can be proved by squaring both sides and fatorization which gives the equivalent $b^2(b-4)^2 \ge 0$. Now proceed like exactly same with that of tchebytchev's argument above.

This problem was proposed by me