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# Inequality 48 (George Basdekis)

Problem:

If $\displaystyle a,b,c,d,e,f\geq 0$ such that $\displaystyle a+b+c+d+e+f=6$ and $\displaystyle a^2+b^2+c^2+d^2+e^2+f^2=\frac{36}{5}$, the prove that

$\displaystyle a^3+b^3+c^3+d^3+e^3+f^3\leq \frac{264}{25}$.

Solution:

Using the Cauchy – Schwarz Inequality we get that

$\displaystyle \frac{36}{5}=a^2+(b^2+...+f^2)\geq a^2+\frac{(b+...+f)^2}{5}=a^2+\frac{(6-a)^2}{5}=\frac{36}{5}+\frac{6a^2-12a}{5}$.

From the above Inequality we can see that $\displaystyle a\leq 2$. Similarly we acquire $\displaystyle b,...,f\leq 2$.

Observe now that

$\displaystyle \sum_{cyc} a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a)$.

Using once again the Cauchy – Schwarz Inequality we get that

$\displaystyle \sum_{cyc} a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}$.

Using the hypothesis, we can calculate the last fraction. Thus we have that

$\displaystyle \sum_{cyc}a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}=\frac{96}{25}$.

Returning to previous Inequality, we now can see that

$\displaystyle \sum_{cyc}a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a)\leq 2\cdot \frac{36}{5}-\frac{96}{25}=\frac{264}{25}$, Q.E.D.