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# Inequality 9(Christos Patilas)

Problem:

If $\displaystyle a_{1},a_{2},a_{3}$ are the positive real roots of the equation $\displaystyle 4x^3-kx^2+mx-9=0$ prove that

$\displaystyle k\geq 4\sqrt[3]{\sum_{cyc}\left(a_{1}\sqrt{a_{2}+a_{3}}\right)+3\prod_{cyc}(a_{1}+a_{2})}$.

Solution:

Let us divide both sides by $\displaystyle 4$ and then cube them.

We acquire

$\displaystyle \left(\frac{k}{4}\right)^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$.

But from Viete’s relations we have that

$\displaystyle \frac{k}{4}=a_{1}+a_{2}+a_{3}$ and $\displaystyle a_{1}a_{2}a_{3}=\frac{9}{4}$.

So our inequality transforms into

$\displaystyle (a_{1}+a_{2}+a_{3})^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$,

or

$\displaystyle \sum_{cyc} a^{3}_{1}+3\prod_{cyc}(a_{1}+a_{2}) \geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$.

So, it suffices to prove that

$\displaystyle \sum_{cyc} a^{3}_{1}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}$.

But the last inequality holds because

$\displaystyle a^{3}_{1}+a^{3}_{2}+a^{3}_{3}\geq a_{1}a_{2}(a_{1}+a_{2})+a^{3}_{3}\geq 2\sqrt{a_{1}a_{2}a_{3}\cdot a^{2}_{3}(a_{1}+a_{2})}=3a_{3}\sqrt{a_{1}+a_{2}}$.

Adding up the $\displaystyle 3$ cyclic relations we come to the desired inequality, Q.E.D.