# An easy inequality?

While solving a problem, I have struck over this inequality, as

If $a^3+b^3+c^3=15$, find minimum value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ Can anybody help me?

• – Ant Sep 25 '14 at 9:38
• Hem, where is the inequality ? And what is the domain of the variables ? – Yves Daoust Sep 25 '14 at 9:41
• Given the symmetry of the problem, and the fact that all the operations involved ($x\to x^3$, $x\to1/x$, $(x,y,z)\to x+y+z$) are fairly simple and monotone, one would generally expect the minimum to be either on the symmetry axis (i.e. where $a=b=c=\sqrt[3]{15/3}$) or as far away from it as possible (i.e. at $a=b=0$, $c=\sqrt[3]{15}$, assuming that all three variables must be non-negative). A natural approach, then, would be to try both, see which one yields the smaller sum of reciprocals, and then see if you can prove that the sum cannot get smaller than that. – Ilmari Karonen Sep 25 '14 at 13:50

By Holder's Inequality:$$\left(\frac1a+\frac1b+\frac1c \right)^3(a^3+b^3+c^3) \ge (1+1+1)^4$$ $$\implies \frac1a+\frac1b+\frac1c \ge \frac3{\sqrt[3]5}$$ with equality when $a=b=c=\sqrt[3]5$.
• In the above, I have assumed $a, b, c > 0$, if negative reals are allowed, there is no lower bound... – Macavity Sep 25 '14 at 9:50