# A Warmup Inequality from Vasile Cirtoaje

### Problem

### Solution

By the AM-GM inequality,

$\displaystyle bc\le\frac{b^3+c^3+1}{3}=\frac{4-a^3}{3}.$

It follows that

$\displaystyle b^4c^4\le\frac{4b^3c^3-a^3b^3c^3}{3}.$

Similarly,

$\displaystyle\begin{align} &c^4a^4\le\frac{4c^3a^3-a^3b^3c^3}{3}\\ &a^4b^4\le\frac{4a^3b^3-a^3b^3c^3}{3}. \end{align}$

Adding all three gives

$\displaystyle \sum_{cycl}b^4c^4\le\frac{4}{3}\sum_{cycl}b^3c^3-a^3b^3c^3.$

Thus, suffice it to show that

$\displaystyle 4\sum_{cycl}b^3c^3-3a^3b^3c^3\le 9$

which is just the third degree Schur's inequality

$4(xy+yz+zx)(x+y+z)-9xyz\le (x+y+z)^2$

for $x=a^3, y=b^2, z=c^2.$ Equality occurs for $a=b=c=1.$

### Acknowledgment

This is problem 1.4 from Vasile Cirtoaje book Algebraic Inequalities (GIL Publishing House, 2006). The solution above is by Gabriel Dospinescu.

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