parmenides51 wrote: Let $a,b$ and $c$ be positive numbers with $a^2+b^2+c^2=3$. Prove that $$a+b+c\ge 3\sqrt[5]{abc}.$$ Solution of Zhangyanzong: $$3^5abc(a+b+c)=3^4abc(a+b+c)(a^2+b^2+c^2)\leq 3^3(ab+bc+ca)^2(a^2+b^2+c^2)$$$$\leq 3^3\left(\frac{2(ab+bc+ca)+a^2+b^2+c^2}{3}\right)^3=(a+b+c)^6,$$$$a+b+c\ge 3\sqrt[5]{abc}.$$