Home » Uncategorized » Inequality 48 (George Basdekis)

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.

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